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The coefficients of x^{p} and x^{q} in the expansion of (1+ x)^{p+q} are[2002]
We have t_{p + 1 }= ^{p + q}C_{p} x^{p} and t_{q + 1} = ^{p+q}C_{q }x^{q}
^{p + q}C_{p} = ^{p + q}C_{q}.
[ Remember ^{ n}C_{r }= ^{n}C_{n – r }]
If the sum of the coefficients in the expansion of ( a + b)^{n} is 4096, then the greatest coeficient in the expansion is [2002]
We have 2^{n} = 4096 = 2^{12} ⇒ n = 12; the greatest coefff = coeff of middle term. So middle term
= t_{7}.; t_{7} = t_{6 + 1} ⇒ coeff of
The positive integer just greater than (1 + 0.0001)10000 is [2002]
(1 + 0.0001)^{10000} n = 10000
r and n are positive integers r > 1, n > 2 and coefficient of (r+2)^{th} term and 3r^{th }term in the expansion of (1 + x)^{2n} are equal, then n equals [2002]
t_{r + 2} = ^{2n}C_{r} + 1 x^{r + 1};t_{3r} = ^{2n}C_{3r – 1} x^{3r – 1 }
Given^{ 2n}C_{r + 1} = ^{2n}C_{3r – 1 };
⇒ ^{2n}C_{2n – (r + 1)} = ^{2n}C_{3r – 1}
⇒ 2n – r – 1 = 3r – 1
⇒ 2n = 4r
⇒ n = 2r
If haing n radical signs then by methods of mathematical induction which is true [2002]
Then = 7 + a_{m} < 7 + 7 < 14.
So by the prin ciple of mathematical induction a_{n} < 7 ∀ n.
If x is positive, the first negative term in the expansion of (1 + x)^{27/5} is [2003]
For first negative term, n  r + 1 < 0 ⇒ r >n+1
Therefore, first negative term is T8 .
The number of integral terms in the expansion of is [2003]
Terms will be integral if both are +ve
integer, which is so if r is an integral multiple of 8. As 0 ≤ r ≤ 256
∴ r = 0,8,16,24,........256 , total 33 values.
Let S(K) = 1 + 3 + 5... + (2K  1) =3+K^{ 2} . Then which of the following is true [2004]
S(k) = 1+3+5+...+(2k – 1) = 3 + k^{2}
S (1) : 1 = 3+ 1, which is not true
∵ S (1) is not true.
∴ P.M.I cannot be applied Let S(k) is true, i.e.
1 + 3 + 5.... + (2k  1) =3+k^{2 }
⇒ 1 + 3 + 5....+ (2k1) + 2k +1
= 3+ k^{2} + 2k +1 = 3+ (k+1)^{2}
∴ S (k ) ⇒ S (k+ 1)
The coefficient of the middle term in the binomial expansion in powers of x of (1 + αx)^{4} and of (1  αx)^{6} is the same if a equals [2004]
The middle term in the expansion of
(1 + αx)^{4} = T_{3} = ^{4}C_{2} (αx)^{2} =6α^{2}x^{2}
The middle term in the expansion of
(1  αx)^{6} = T_{4} = ^{6}C_{3} ( αx)^{3} = 20α^{2}x^{2}
According to the question
6α^{ 2} = 20α^{3} ⇒ α = 
The coefficient of xn in expansion of (1 + x) (1 – x)^{n} is
Coeff of x^{n} in (1 + x)(1x )^{n }
= Coeff of x^{n} in (1  x)^{n}+ Coeff of x^{n 1 }in (1 x )_{n}
= ( 1)^{n} ^{n}C_{n} + (1)^{n} ^{1 n}C_{n1 }
= ( 1)^{n} 1 + (1)^{n 1} n
= ( 1)^{n} [1 n]
The value of [2005]
We know ⇒ (^{50 }C_{4 }+^{50}C_{3})
+ ^{51}C_{3 }+ ^{52 }C_{3} + ^{53}C_{3} + ^{54 }C_{3}+ ^{55}C_{3}
⇒ (^{51} C_{4} + ^{51}C_{3}) + ^{52} C_{3} + ^{53}C_{3} + ^{54} C_{3}+^{ 55}C_{3}
Proceeding in the same way, we get
⇒ ^{54}C_{4} + ^{55}C_{3} = ^{56}C_{4} .
IF then which one of the following holds for all n ≥ 1, by the principle of mathematical induction [2005]
We observe that
and we can prove by
induction that
Now nA  ( n 1)I =
∴ nA  (n  1) I = A^{n}
If the coefficient of x^{ 7} in equals the coefficient of x^{ 7} in , then a and b satisfy the relation [2005]
T_{r +1} in the expansion
For the Coefficient of x^{7}, we have
⇒ 22 – 3r = 7 ⇒ r = 5
∴ Coefficient of x^{7 }= ^{11}C_{5} (a)^{6} (b)^{ 5} ...(1)
Again T_{r + 1} in the expansion
= ^{11}C_{r }(a)^{11 r }( 1)^{r }_{x} (b)^{ r} ( x)^{ 2r} (x)^{11r}
For the Coefficient of x^{–7}, we have
Now 11 – 3r = – 7 ⇒ 3r = 18 ⇒ r = 6
∴ Coefficient of x^{ 7 }= ^{11}C_{6} a^{5 }x 1 x (b)^{6 }
∴ Coefficient of x^{7 }= Coefficient of x^{–7 }
⇒ = ^{11}C_{6} (a)^{6 }x (b)^{5 } = ^{11}C_{6} a^{5 }x (b)^{6 }⇒ ab = 1.
If x is so small that x^{3} and higher powers of x may be neglected, then may be approximated as [2005]
∵ x^{3 }and higher powers of x may be neglected
(as x^{3} and higher powers of x can be neglected)
If the expansion in powers of x of the function is a_{0} + a_{1} x + a_{2} x^{2}+ a_{3}x^{3} ...... then a_{n} is [2006]
(1  ax)^{ 1} (1 bx)^{1}
= (1 + ax + a^{ 2}x^{2 }+ ...)(1 + bx + b^{ 2}x^{2}+ ...)
∴ Coefficient of x^{n}
x^{n} = b^{n} + ab^{n 1} + a^{2}b^{n  2} + ....... + a^{n1}b+a^{n}
{which is a G.P. with
∴ Its sum is =
For natural numbers m, n if (1 – y)^{m} (1 + y)^{n} = 1 + a_{1}y + a_{2}y^{2} + ....... and a_{1} = a_{2} = 10, then (m, n) is
(1  y) ^{m} (1+y)^{n}
= 1 + (nm) +
∴ a_{1} = n m = 10
and
So, n – m = 10 and (m n)^{2}  (m +n)= 20
⇒ m +n= 80
∴ m = 35, n = 45
In the binomial expansion of (a – b)^{n}, n ≥ 5, the sum of 5^{th} and 6^{th} terms is zero, then a/b equals
T_{r + 1} = (–1)^{r}. ^{n}C_{r} (a)^{n – r}. (b)^{r} is an expansion of (a – b)^{n}
∴ 5th term = t_{5} = t_{4+1}
= (–1)^{4}.^{ n}C_{4} (a)^{n–4}.(b)^{4} = ^{n}C_{4} . a^{n–4} . b^{4 }
6th term = t_{6 }= t_{5+1} = (–1)^{5 n}C_{5} (a)^{n–5} (b)^{5 }
Given t_{5} + t_{6} = 0
∴ ^{n}C_{4 }. a^{n–4} . b^{4} + (– ^{n}C_{5} . a^{n–5} . b^{5}) = 0
or,
The sum of the series is [2007]
We know that, (1 + x)^{20 }= ^{20}C_{0} + ^{20}C_{1}x + ^{20}C_{2} x^{2} + ...... ^{20}C_{10 }x^{10 }+ ..... ^{20}C_{20 }x^{20}
Put x = –1, (0) = ^{20}C_{0} – ^{20}C_{1} +^{ 20}C_{2} – ^{20}C_{3} + ...... + ^{20}C_{10 }– ^{20}C_{11} .... + ^{20}C_{20}
⇒ 0 = 2[^{20}C_{0} – ^{20}C_{1} + ^{20}C_{2} – ^{20}C_{3}
+ ..... – ^{20}C_{9}] + ^{20}C_{10}
⇒^{ 20}C_{10 }= 2[^{20}C_{0 }– ^{20}C_{1 }+ ^{20}C_{2} – ^{20}C_{3}
+ ...... – ^{20}C_{9} + ^{20}C_{10}]
⇒ ^{20}C_{0 }– ^{20}C_{1} + ^{20}C_{2} – ^{20}C_{3 }+ .... +^{ 20}C_{10} = ^{20}C_{10}
Statement 1 : [2008]
Statement2 :
We have
= nx (1+ x)^{ n–1} + (1+ x)^{n} = RHS
∴ Statement 2 is correct.
Putting x = 1, we get
∴ Statement 1 is also true and statement 2 is a correct explanation for statement 1.
The remainder left out when 8^{2n }– (62)^{2n +1} is divided by 9 is :[2009]
(8)^{2n} – (62) ^{2n + 1 }= (64) ^{n }– (62)^{2n + 1}
= (63 + 1)^{n} – (63 – 1)^{2n + 1}
= 63 × +........]+ 1
– 63 ×
⇒ 63 × some integral value + 2
⇒ 8^{2n }– (62)^{2n+1 }when divided by 9 leaves 2 as the remainder.
Let
Statement1 : S_{3} = 55 × 2^{9}
Statement2: S_{1} = 90 × 2^{8} and S_{2 }= 10 × 2^{8 }. [2010]
The coefficient of x^{7} in the expansion of (1– x – x^{2} + x^{3} )^{6} is [2011]
(1 – x – x^{2 }+ x^{3})^{6} = [(1– x) – x^{2 }(1 – x)]^{6 }
= (1– x)^{6} (1 – x^{2})^{6 }
= (1 – 6x + 15x^{2 }– 20x^{3} + 15x^{4} – 6x^{5} + x^{6})
× (1 – 6x^{2} + 15x^{4 }– 20x^{6 }+ 15x^{8} – 6x^{10} + x^{12})
Coefficient of x^{7} = (– 6) (– 20) + (– 20)(15) + (– 6) (–6)
= – 144
If n is a positive integer , then is
= x Some integer ∴ irrational number
The term independent of x in expansion of is [JEE M 2013]
Given expression can be written as
= (x^{1/3} – x^{–1/2})^{10}
General term = T_{r+1= }^{10}C_{r}_{ }(x^{1/3}_{)}^{10–r}(–x^{–1/2})^{r}
Term will be independent of x when
⇒ r = 4
So, required term = T_{5} = ^{10}C_{4} = 210
If the coefficents of x^{3} and x^{4} in the expansion of (1 + ax + bx^{2} ) (12 x )^{18} in powers of x are both zero, then (a, b) is equal to: [JEE M 2014]
Consider (1 + ax + bx^{2}) (1 – 2x)^{18}
= (1 + ax + bx^{2}) [^{18}C_{0 }– ^{18}C_{1} (2x) + ^{18}C_{2}(2x)^{2}– ^{18}C_{3}(2x)^{3 }+ ^{18}C_{4}(2x)^{4} –.......]
Coeff of x^{3 }=^{ 18}C_{3} (–2)^{3} + a. (–2)^{2.} ^{18}C_{2} + b (–2). ^{18}C_{1 }= 0 Coeff. of x^{3} = – ^{18}C_{3}.8 + a × 4. ^{18}C_{2}– 2b × 18 = 0
= –51 × 16 × 8 + a × 36 × 17 – 36b = 0
= –34 × 16 + 51a – 3b = 0
= 51a – 3b = 34 × 16 = 544
= 51a – 3b = 544 ....(i)
Only option number (b) satisfies the equation number (i)
The sum of coefficients of integral power of x in the binomial expansion is: [JEE M 2015]
...(1)
...(2)
Adding equation (1) and (2)
Putting x = 1, we get above as
If the number of terms in th e expan sion of x ≠ 0, is 28, then the sum of the coefficients of all the terms in this expansion, is : [JEE M 2016]
Total number of terms = ^{n+2}C_{2} = 28 (n + 2) (n + 1) = 56
x = 6
Sum of coefficients = (1 – 2 + 4)^{n} = 3^{6} = 729
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