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All questions of Binomial Theorem for Commerce Exam

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The expansion of  , in powers of x, is valid if
  • a)
    |x| > 2
  • b)
    x < 2
  • c)
    x > 2
  • d)
    |x| < 2
Correct answer is option 'D'. Can you explain this answer?

Knowledge Hub answered
In case of negative or fractional power, expansion (1+x)^n is valid only when |x| < 1
(6 - 3x)-1/2
= (6-1/2 (1 - x/2)-1/2)
So, this equation exists only when |x/2| < 1
|x| < 2

If n is a +ve integer, then the binomial coefficients equidistant from the beginning and the end in the expansion of (x+a)n are
  • a)
    additive inverse of each other
  • b)
    multiplicative inverse of each other
  • c)
    equal
  • d)
    nothing can be said
Correct answer is option 'C'. Can you explain this answer?

Hansa Sharma answered
(x+a)n = nC0 xn + nC1 x(n-1) a1 + nC2 x(n-2) a2 + ..........+ nC(n-1) xa(n-1) + nCn  an
Now, nC0 = nCn, nC1 = nCn-1,    nC2 = nCn-2,........
therefore, nCr = nCn-r
The binomial coefficients equidistant from the beginning and the end in the expansion of (x+a)n are equal.

The value of 1261/3 upto three decimals is
  • a)
    5.013
  • b)
    5.014
  • c)
    5.012
  • d)
    5.011
Correct answer is option 'A'. Can you explain this answer?

Lavanya Menon answered
(126)11/3
=  (125 + 1)1/3
=  [125 (1 + (1/125))]1/3
=  1251/3(1 + (1/125))1/3         1/125 < 1
= 5 [1 + (1/3)(1/125) + ..........]
= 5 [1 + (1/3)(0.008)]
= 5 [1 + 0.002666]
= 5.013

the coefficient of xn in the expansion of 
  • a)
    n
  • b)
    2n
  • c)
    4n
  • d)
    zero
Correct answer is option 'C'. Can you explain this answer?

Suresh Reddy answered
[(1+x)/(1−x)]2
⇒ (1+x)2(1−x)-2
⇒ (1 + x2 + 2x)[1 + 2x + 3x2 +..... +(n−1)xn-2 + nxn-1 + (n+1)xn +...]
coeff of xn will be given by
(I) When 1 will be multiplied by (n+1)xn
(II) When x2 will be multiplied by (n−1)xn-1
(III) When 2x will be multiplied by nxn-1
∴ coeff. of xn = n + 1 + n − 1 + 2n
= 4n

The coefficient of y in the expansion of (y² + c/y)5 is 
  • a)
    10c 
  • b)
    10c² 
  • c)
    10c³ 
  • d)
    None of these
Correct answer is option 'C'. Can you explain this answer?

Varun Kapoor answered
Given, binomial expression is (y² + c / y)5 
Now, Tr+1 = 5Cr × (y²)5-r × (c / y)r 
= 5Cr × y10-3r × Cr 
Now, 10 – 3r = 1 
⇒ 3r = 9 
⇒ r = 3 
So, the coefficient of y = 5C3 × c³ = 10c³

 What is the general term in the expansion of (2y-4x)44?
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'A'. Can you explain this answer?

Om Desai answered
 n = 44, p = 2y q = -4x
General term of (p+q)n is given by
T(r+1) = nCr . pr . q(n-r)
= 44Cr . (2y)r . (-4x)(44-r)

The coefficient of x4 in the expansion of (1 + x + x2 + x3)n is:
  • a)
    nC4 + nC2
  • b)
    nC4 + nC2+ nC4.nC2
  • c)
    nC4 + nC2+ nC1.nC2
  • d)
    nC4
Correct answer is option 'C'. Can you explain this answer?

Gaurav Kumar answered
x4 can be achieved in the following ways: 
x4 . 1(n-4) . (x2)0 . (x3)0
Hence, coefficient will be  nC4 .
x2 . 1(n-3) . (x2)1 . (x3)0
Hence, coefficient will be 3nC3.
x1 . 1(n-2) . (x2)0 . (x3)1
Hence, coefficient will be 2nC2.
x0 . 1(n-2) .(x2)2 .(x3)0
Hence, coefficient will be nC2 .
Hence, the required coefficient will be 
nC4 + 3nC3 + 3nC2
= nC4 + 3(nC3 + nC2).
= nC4  + 3(n+1C3).
= nC4  + nC2 + nC1 . nC2

Which of the following is divisible by 25:
  • a)
    6n - 5n + 1
  • b)
    6n + 5n
  • c)
    6n - 5n
  • d)
    6n - 5n - 1
Correct answer is option 'D'. Can you explain this answer?

Riya Banerjee answered
we can write (6ⁿ ) = (1 + 5)ⁿ
we know, according to binomial theorem,
(1 + x)ⁿ = 1 + nx + n(n-1)x²/2! + n(n-1)(n-2)x³/3! +.............∞ use this here,
(6)ⁿ = (1 + 5)ⁿ = 1 + 5n + n(n-1)5²/2! + n(n-1)(n-2)5³/3! +...........∞
= 1 + 5n + 5²{ n(n-1)/2! + n(n-1)(n-2)5/3! +.......∞}
Let P = n(n-1)/2! + n(n-1)(n-2)5/3! +.........∞
6ⁿ = 1 + 5n + 25P
6ⁿ - 5n = 1 + 25P -------(1)
but we know, according to Euclid algorithm ,
dividend = divisor × quotient + remainder ---(2)
compare eqn (1) to (2)
we observed that 6ⁿ -5 n always leaves the remainder 1 when divided by 25

If 2nd, 3rd and 4th terms in the expansion of (x+a)n are 240, 720 and 1080 respectively, then the value of n is
  • a)
    15
  • b)
    20
  • c)
    5
  • d)
    10
Correct answer is option 'C'. Can you explain this answer?

Shiksha Academy answered
General term Tr+1 of (x+y)n is given by 
Tr+1 = nCr xn-r yr
T2 = nC2 xn-2 y = 240
T3 = nC3 xn-3 y2 = 720
T4 = nC4 xn-4 y3 = 1080
T3/T2 = [(n-1)/2] * [y/x] = 3......(1)
T4/T2 = {[(n-1)(n-2)]/(3*2)} * x2/y2 = 9/2
T4/T3 = [(n-2)/3] * [y/x] = 3/2...(2)
Dividing 1 by 2
[(n-1)/2] * [3/(n-2)] = 2
⇒ 3n−3 = 4n−8
⇒ 5 = n

The coefficient of x3 in the binomial expansion of   
  • a)
    792m5
  • b)
    942m7
  • c)
    330m4
  • d)
    792m6
Correct answer is option 'C'. Can you explain this answer?

Geetika Shah answered
T(r+1) = 11Cr x(11-2r) (m/r)r (-1)r
= 11Cr x(11-2r) mr (-1)r
Coefficient of x3 = 11 - 2r = 3
8 = 2r 
r = 4
T5 = 11C4 x3 m4 (-1)
Coefficient of x3 = 11C4 m4
= 330 m4

The coefficient of x17 in the expansion of (x- 1) (x- 2) …..(x – 18) is
  • a)
    - 171
  • b)
    342
  • c)
    171/2
  • d)
    684
Correct answer is option 'A'. Can you explain this answer?

Infinity Academy answered
The coefficient of x17 is given by 
−1 + (−2) + (−3) + ….. (−18)
= −1 − 2 − 3….. − 18
= − (18(18+1))/2
= − 9(19)
= − 171

In the expansion of (1+x)60, the sum of coefficients of odd powers of x is
  • a)
    261
  • b)
    260
  • c)
    2
    59
  • d)
    none of these.
Correct answer is option 'C'. Can you explain this answer?

Mohit Mittal answered
C is correct as
sum of ( 1+ x )^60 gives 2^60 which includes both even and odd powers of x
so for only one type of power ( odd power) of x we divide 2^ 60 by 2 so we get 2^ 59

The middle term in the expansion of (2x+3y)12 is
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'B'. Can you explain this answer?

Arun Khanna answered
There will be 13 terms, so the middle term is term #7
Term(7) = C(12,6)(2x)^6 (3y)^6
= 924(64x^6)(729y^6)
= 43110144 x^6 y^6
so the correct option is B

The coefficient of second, third and fourth terms in the binomial expansion of (1+x)n(‘n’, a + ve integer) are in A.P.., if n is equal to
  • a)
    4
  • b)
    7
  • c)
    5
  • d)
    6
Correct answer is option 'B'. Can you explain this answer?

Pallabi Patel answered
Explanation:

To find the coefficient of the second, third, and fourth terms in the binomial expansion of (1+x)^n, we need to understand the general formula for the binomial expansion and the concept of the arithmetic progression (A.P.).

Binomial Expansion:
The binomial expansion of (1+x)^n can be calculated using the binomial theorem. According to the binomial theorem, the expansion is given by:

(1+x)^n = C(n,0) + C(n,1)x + C(n,2)x^2 + C(n,3)x^3 + ... + C(n,n)x^n

where C(n,r) represents the binomial coefficient, given by:

C(n,r) = n! / (r!(n-r)!)

Arithmetic Progression (A.P.):
An arithmetic progression (A.P.) is a sequence of numbers in which the difference between any two consecutive terms is constant. The common difference between the terms is denoted by 'd'.

For an A.P., the nth term (Tn) can be calculated using the formula:
Tn = a + (n-1)d

where 'a' is the first term and 'd' is the common difference.

Coeficients in A.P.:
Now, to determine the coefficients of the second, third, and fourth terms in the binomial expansion, we need to find the values of r for which the binomial coefficients form an arithmetic progression.

From the binomial expansion formula, we can see that the coefficients C(n,1), C(n,2), and C(n,3) correspond to the second, third, and fourth terms respectively.

Let's calculate the difference between the consecutive binomial coefficients:

d1 = C(n,2) - C(n,1) = (n! / (2!(n-2)!)) - (n! / (1!(n-1)!))
= (n(n-1)(n-2)! / (2(n-2)!)) - (n(n-1)(n-2)! / (1(n-1)!))
= n(n-1)(n-2)! / (2(n-2)!) - n(n-1)(n-2)! / (1(n-1)!)
= n(n-1) / 2 - n(n-1)
= -n(n-1) / 2

d2 = C(n,3) - C(n,2) = (n! / (3!(n-3)!)) - (n! / (2!(n-2)!))
= (n(n-1)(n-2)! / (3(n-3)!)) - (n(n-1)(n-2)! / (2(n-2)!))
= n(n-1) / 3 - n(n-1) / 2
= -n(n-1) / 6

We can observe that the differences d1 and d2 are both negative and have a common factor of -n(n-1). This implies that the coefficients C(n,1), C(n,2), and C(n,3) form an arithmetic progression with a common difference of -n(n-1)/2.

Conclusion:
Since the coefficients of the second,

If the second, third, and fourth terms in the expansion of (x + y)n are 135, 30, and 10/3, respectively, then 6 (n3 + x2 + y) is equal to ________.
Correct answer is '806'. Can you explain this answer?

Nipuns Institute answered
T2 = nC1 y1 x(n-1) = 135
T3 = nC2 y2 x(n-2) = 30
T4 = nC3 y3 x(n-3) = 10/3
⇒ 135/30 = (x/y) * n * 2 / n(n-1) = (2 / n-1) * (x/y) ... (i)
30 / (10/3) = n(n-1) / 2 / n(n-1)(n-2) * 3! * (x/y)
9 = (3 / n-2) * (x/y)
3(n - 2) = 135 / 60 (n - 1) ⇒ n = 5
⇒ x = 9y ... (i)
y * x4 = 27 ⇒ x / 9 * x4 = 33
⇒ x5 = 35 ⇒ x = 3y = 1/3
⇒ 6 (53 + 32 + 1/3) = 6 (125 + 9 + 1/3)
= 6(134) + 2 = 806

Number of integral terms in the expansion of (71/2 + 111/6)824 is equal to ________.
Correct answer is '138'. Can you explain this answer?

Sai Kulkarni answered
General term in expansion of (71/2 + 111/6)824 is T(r+1) = 824Cr * 7((824-r)/2) * 11(r/6)
For integral term, r must be a multiple of 6.
Hence r = 0, 6, 12, ..., 822

n-1Cr = (k² - 8) nCr+1 if and only if:
  • a)
    2√2 < k < 2√3
  • b)
    2√2 ≤ k ≤ 3
  • c)
    2√3 < k < 3√3
  • d)
    2√3 ≤ k ≤ 3√2
Correct answer is option 'B'. Can you explain this answer?

Preethi Bose answered
Understanding the Given Equation
The equation provided is n-1Cr = (k² - 8) nCr+1. To analyze the conditions for k, we need to explore the relationship between the coefficients of binomial expansions.
Analyzing n-1Cr and nCr+1
- n-1Cr represents the number of ways to choose r items from n-1 items.
- nCr+1 represents the number of ways to choose r+1 items from n items.
The equation can be interpreted as establishing a proportional relationship between the two coefficients based on the value of k.
Rearranging the Equation
1. Rewrite the equation: k² - 8 = (n-1Cr) / (nCr+1).
2. The left side, k² - 8, must yield a non-negative value for valid k.
Finding the Range of k
- The expression (n-1Cr)/(nCr+1) must lie within certain bounds based on the properties of binomial coefficients.
- As n increases, the ratio approaches 1, thus k² must be greater than or equal to 8, leading to k ≥ 2√2.
- Further analysis of the equation indicates an upper bound leading to k ≤ 3.
Conclusion: Valid Range for k
- The derived inequalities give 2√2 ≤ k ≤ 3.
- This range confirms that option B (2√2 ≤ k ≤ 3) is the correct answer, ensuring the equation holds true under the specified conditions.
In conclusion, the relationship between the coefficients provides a clear range for k, validating option B as the correct choice for the equation provided.

The greatest coefficient in the expansion of (1+x)12 is
  • a)
    C (12, 4)
  • b)
    C (12, 6)
  • c)
    C (12, 5)
  • d)
    none of these
Correct answer is option 'B'. Can you explain this answer?

Harshad Unni answered
Explanation:
To find the greatest coefficient in the expansion of (1 + x)^12, we can use the binomial theorem. The binomial theorem states that for any positive integer n, the expansion of (a + b)^n can be written as:

(a + b)^n = C(n, 0) * a^n * b^0 + C(n, 1) * a^(n-1) * b^1 + C(n, 2) * a^(n-2) * b^2 + ... + C(n, n-1) * a^1 * b^(n-1) + C(n, n) * a^0 * b^n

where C(n, k) represents the binomial coefficient, also known as "n choose k".

In this case, we have (1 + x)^12, so a = 1 and b = x. Plugging these values into the binomial theorem formula, we get:

(1 + x)^12 = C(12, 0) * 1^12 * x^0 + C(12, 1) * 1^11 * x^1 + C(12, 2) * 1^10 * x^2 + ... + C(12, 11) * 1^1 * x^11 + C(12, 12) * 1^0 * x^12

Simplifying this expression, we have:

(1 + x)^12 = C(12, 0) + C(12, 1) * x + C(12, 2) * x^2 + ... + C(12, 11) * x^11 + C(12, 12) * x^12

To find the greatest coefficient, we need to determine which term has the highest coefficient. The coefficient of each term is given by the binomial coefficient C(12, k), where k represents the power of x.

Comparing Coefficients:
In order to find the greatest coefficient, we need to compare the binomial coefficients for each term. The binomial coefficient C(12, k) can be calculated using the formula:

C(n, k) = n! / (k! * (n - k)!)

where "!" represents the factorial operation.

Let's calculate the binomial coefficients for the given options:

a) C(12, 4) = 12! / (4! * (12 - 4)!) = 12! / (4! * 8!) = 495
b) C(12, 6) = 12! / (6! * (12 - 6)!) = 12! / (6! * 6!) = 924
c) C(12, 5) = 12! / (5! * (12 - 5)!) = 12! / (5! * 7!) = 792

Comparing the coefficients, we can see that option B has the greatest coefficient, which is 924. Therefore, the correct answer is option B) C(12, 6).

If x = 9950+1005 and y = (101)50, then
  • a)
    x < y
  • b)
    x = y
  • c)
    x > y
  • d)
    none of these
Correct answer is option 'A'. Can you explain this answer?

Nitya Yadav answered
The given equations are:
Ifx = 9950
1005andy = (101)50

To solve for x, we divide both sides of the first equation by 100:
Ifx/100 = 9950/100
x = 99.5

To solve for y, we divide both sides of the second equation by 1005:
andy = (101)50/1005
andy = 50

Therefore, the values of x and y are:
x = 99.5
y = 50

The coefficient of xn in expansion of (1+x)(1−x)n is
  • a)
    (−1)n(n+1)
  • b)
    (−1)n(1−n)
  • c)
    (−1)n
  • d)
    none of these
Correct answer is option 'B'. Can you explain this answer?

Lavanya Menon answered
We can expand (1+x)(1−x)n as 
= (1+x)(nC0​− nC1​.x + nC2​.x2 + ........ +(−1)n nCn​xn)

Coefficient of xn is 
(−1)n nCn ​+ (−1)n−1 nCn−1
​=(−1)n(1−n)

If in the expansion of(1+x)43, the coefficients of (2r+1)th and (r+2)th terms are equal, then r is equal to
  • a)
    7 or 1
  • b)
    21 or 1
  • c)
    1 or 14
  • d)
    none of these
Correct answer is option 'C'. Can you explain this answer?

Pallabi Nambiar answered
Understanding the Problem
In the expansion of (1 + x)^43, we need to find the values of r for which the coefficients of the (2r + 1)th term and the (r + 2)th term are equal.
Finding the Coefficients
The general term T_k in the expansion of (1 + x)^n can be expressed as:
- T_k = C(n, k) * x^k
Where C(n, k) is the binomial coefficient.
For our case, n = 43:
- Coefficient of the (2r + 1)th term: C(43, 2r)
- Coefficient of the (r + 2)th term: C(43, r)
We set these coefficients equal:
- C(43, 2r) = C(43, r)
Using Binomial Coefficient Properties
Using the property of binomial coefficients, we know that:
- C(n, k) = C(n, n - k)
Thus, we can rewrite:
- C(43, 2r) = C(43, 43 - 2r)
- C(43, r) = C(43, 43 - r)
This leads to two scenarios:
1. 2r = r → r = 0 (not valid since we want a positive r)
2. 2r + r = 43 → 3r = 43 → r = 14.33 (not an integer)
Now we consider:
Finding Integer Values for r
Considering the nature of binomial coefficients, we check for values of r:
- r can be integers such that 2r ≤ 43 and r ≤ 43.
- Possible values of r include 1, 7, 14, and others.
By checking:
- If r = 1: C(43, 2) = C(43, 1)
- If r = 14: C(43, 28) = C(43, 14)
Both conditions satisfy the equality.
Conclusion
Thus, the values of r are 1 and 14. Therefore, the correct option is:
- Option C: 1 or 14.

If rth ,(r+1)th and (r+2)th terms in the expansion of (1+x)n are in A.P. then
  • a)
    (n+2r)3 = n−2
  • b)
    (n−2r)2 = n+2
  • c)
    (n+2r)2 = n+2
  • d)
    none of these
Correct answer is option 'B'. Can you explain this answer?

Mayank Dasgupta answered
To determine the relationship between the terms (rth, (r+1)th, and (r+2)th) in the expansion of (1+x)^n, we can use the binomial theorem.

The general term of the expansion of (1+x)^n is given by:
C(n, r)*x^r*(1)^n-r

The rth term is given by:
C(n, r)*x^r*(1)^(n-r)

The (r+1)th term is given by:
C(n, r+1)*x^(r+1)*(1)^(n-r-1)

The (r+2)th term is given by:
C(n, r+2)*x^(r+2)*(1)^(n-r-2)

To determine if these terms are in an arithmetic progression (A.P.), we need to find the common difference between the terms. We can subtract consecutive terms to find the common difference:

(r+1)th term - rth term:
[C(n, r+1)*x^(r+1)*(1)^(n-r-1)] - [C(n, r)*x^r*(1)^(n-r)]
Simplifying this expression, we get:
C(n, r+1)*x^(r+1)*(1)^(n-r-1) - C(n, r)*x^r*(1)^(n-r)
= [C(n, r+1)*x^(r+1)*(1)^(n-r-1) - C(n, r)*x^r*(1)^(n-r)] / [x^r*(1)^(n-r)]

Similarly, the (r+2)th term - (r+1)th term can be found as:
[C(n, r+2)*x^(r+2)*(1)^(n-r-2)] - [C(n, r+1)*x^(r+1)*(1)^(n-r-1)]
Simplifying this expression, we get:
C(n, r+2)*x^(r+2)*(1)^(n-r-2) - C(n, r+1)*x^(r+1)*(1)^(n-r-1)
= [C(n, r+2)*x^(r+2)*(1)^(n-r-2) - C(n, r+1)*x^(r+1)*(1)^(n-r-1)] / [x^(r+1)*(1)^(n-r-1)]

Since we want the terms to be in an arithmetic progression, the common difference between these terms should be the same. Therefore, we can equate the expressions for the common difference and solve for x:

[C(n, r+1)*x^(r+1)*(1)^(n-r-1) - C(n, r)*x^r*(1)^(n-r)] / [x^r*(1)^(n-r)]
= [C(n, r+2)*x^(r+2)*(1)^(n-r-2) - C(n, r+1)*x^(r+1)*(1)^(n-r-1)] / [x^(r+1)*(1)^(n-r-1)]

Simplifying this expression and canceling out the common factors, we get:
[C(n, r+1)*x -

The sum of coefficients in the expansion of (x+2y+z)n is (n being a positive integer)
  • a)
    3n
  • b)
    2n
  • c)
    1
  • d)
    none of these
Correct answer is option 'D'. Can you explain this answer?

Maitri Kumar answered
Explanation:

To find the sum of coefficients in the expansion of (x + 2y + z)^n, we can use the Binomial Theorem.

The Binomial Theorem states that for any positive integer n, the expansion of (x + y)^n can be written as:

(x + y)^n = C(n, 0)x^n y^0 + C(n, 1)x^(n-1) y^1 + C(n, 2)x^(n-2) y^2 + ... + C(n, n-1)x^1 y^(n-1) + C(n, n)x^0 y^n

Where C(n, k) is the binomial coefficient, given by:

C(n, k) = n! / (k!(n-k)!)

In the case of (x + 2y + z)^n, the coefficients will be found by replacing x with 1, y with 2, and z with 1. Therefore, the expansion becomes:

(1 + 2(2y) + 1)^n = (1 + 4y + 1)^n

Now we can apply the Binomial Theorem to find the coefficients. The sum of the coefficients will be the sum of the terms in the expansion.

Applying the Binomial Theorem:

Using the Binomial Theorem, the expansion of (1 + 4y + 1)^n can be written as:

(1 + 4y + 1)^n = C(n, 0)(1)^n (4y)^0 + C(n, 1)(1)^(n-1) (4y)^1 + C(n, 2)(1)^(n-2) (4y)^2 + ... + C(n, n-1)(1)^1 (4y)^(n-1) + C(n, n)(1)^0 (4y)^n

Simplifying this expression, we get:

(1 + 4y + 1)^n = C(n, 0) + 4C(n, 1)y + 6C(n, 2)(4y)^2 + ... + nC(n, n-1)(4y)^(n-1) + C(n, n)(4y)^n

The sum of the coefficients will be the sum of the terms in this expansion.

Calculating the Sum of the Coefficients:

To calculate the sum of the coefficients, we need to sum up all the terms in the expansion.

The sum of the coefficients can be written as:

C(n, 0) + 4C(n, 1) + 6C(n, 2) + ... + nC(n, n-1) + C(n, n)

Each term in this sum is a binomial coefficient multiplied by a constant factor.

Conclusion:

In general, the sum of the coefficients in the expansion of (x + 2y + z)^n is not equal to any of the given options (a, b, c). Therefore, the correct answer is option D, "none of these".

5th term from the end in the expansion of 
  • a)
    7920x4
  • b)
    −7920x4
  • c)
    7920x−4
  • d)
    −7920x−4
Correct answer is option 'C'. Can you explain this answer?

Akanksha Reddy answered
5th term in the expansion of (x2/2 − 2/x2)12 is
Tr+ 1= nCr xr* y(n−r)
T5 = 12C4[(x2/2)4 (-2/x2)8]
= (12! * x8 * 28)/ (4! * 8! *24 * x16)
= 7920x−4

The coefficients of xn in the expansion of (1+2x + 3x2 + ........)1/2 is 
  • a)
    1
  • b)
    n
  • c)
    n+1
  • d)
    -1
Correct answer is option 'A'. Can you explain this answer?

Aarya Rane answered
To find the coefficients of xn in the expansion of (1 + 2x + 3x^2 + ...)^1/2, we can use the binomial theorem. The binomial theorem states that for any real number a and b and any positive integer n, the expansion of (a + b)^n can be written as:

(a + b)^n = C(n, 0) * a^n * b^0 + C(n, 1) * a^(n-1) * b^1 + C(n, 2) * a^(n-2) * b^2 + ... + C(n, n-1) * a^1 * b^(n-1) + C(n, n) * a^0 * b^n

where C(n, r) is the binomial coefficient, also called "n choose r", and is given by the formula:

C(n, r) = n! / (r! * (n-r)!)

In the given expression (1 + 2x + 3x^2 + ...)^1/2, we have a = 1 and b = 2x + 3x^2 + ... Therefore, applying the binomial theorem, the expansion becomes:

(1 + 2x + 3x^2 + ...)^1/2 = C(1/2, 0) * 1^(1/2) * (2x + 3x^2 + ...)^0 + C(1/2, 1) * 1^(1/2 - 1) * (2x + 3x^2 + ...)^1/2 + ...

Simplifying this expression, we find that the coefficient of xn is given by:

C(1/2, n) * 1^(1/2 - n) * (2x + 3x^2 + ...)^n/2

Since C(1/2, n) = 1 for all values of n, and 1^(1/2 - n) = 1, the coefficient of xn simplifies to:

(2x + 3x^2 + ...)^n/2

From this, we can see that the coefficient of xn is simply the expression (2x + 3x^2 + ...)^n/2. Since this expression does not depend on n, the coefficient of xn is constant and equal to 1.

Therefore, the correct answer is option A) 1.

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