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Test: General Terms In Binomial Expansion - Commerce MCQ


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10 Questions MCQ Test Mathematics (Maths) Class 11 - Test: General Terms In Binomial Expansion

Test: General Terms In Binomial Expansion for Commerce 2024 is part of Mathematics (Maths) Class 11 preparation. The Test: General Terms In Binomial Expansion questions and answers have been prepared according to the Commerce exam syllabus.The Test: General Terms In Binomial Expansion MCQs are made for Commerce 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: General Terms In Binomial Expansion below.
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Test: General Terms In Binomial Expansion - Question 1

In the expansion of (x+y)n, the coefficients of 4th and 13th terms are equal, Then the value of n is :

Detailed Solution for Test: General Terms In Binomial Expansion - Question 1

Test: General Terms In Binomial Expansion - Question 2

What is the general term in the expansion of (2y-4x)44?

Detailed Solution for Test: General Terms In Binomial Expansion - Question 2

 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)

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Test: General Terms In Binomial Expansion - Question 3

The coefficient of x4 in the expansion of (1 + x + x2 + x3)n is:

Detailed Solution for Test: General Terms In Binomial Expansion - Question 3

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

Test: General Terms In Binomial Expansion - Question 4

The largest coefficient in the expansion of (1+x)24 is:

Detailed Solution for Test: General Terms In Binomial Expansion - Question 4

 Largest coefficient in the expansion of (1+x)24 
= 24C24/2 
= 24C12

Test: General Terms In Binomial Expansion - Question 5

Which of the following is divisible by 25:

Detailed Solution for Test: General Terms In Binomial Expansion - Question 5

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

Test: General Terms In Binomial Expansion - Question 6

In the expansion of (1+a)m+n which of the following is true?

Detailed Solution for Test: General Terms In Binomial Expansion - Question 6

 (1+a)m+n
Coefficient of am = m+nCm = m+n
Coefficient of an = m+nCn
nCx = nC(n−x) Property
Hence, coefficients of am and an are equal.

Test: General Terms In Binomial Expansion - Question 7

If the third term of the expansion of  is 106 ,then x is equal to 

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Test: General Terms In Binomial Expansion - Question 8

The value of 1261/3 upto three decimals is

Detailed Solution for Test: General Terms In Binomial Expansion - Question 8

(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

Test: General Terms In Binomial Expansion - Question 9

The middle term in the expansion of (2x+3y)12 is

Detailed Solution for Test: General Terms In Binomial Expansion - Question 9

Test: General Terms In Binomial Expansion - Question 10

The coefficient of xn in the expansion of 

Detailed Solution for Test: General Terms In Binomial Expansion - Question 10

[(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

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