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If in the expansion of (1 + x)^{m} (1 – x)^{n}, the coefficients of x and x^{2} are 3 and – 6 respectively, then m is [JEE 99,2 ]
(1+x)^{m} (1−x)^{n} = (1 + mx + m(m−1)x^{2}/2!)(1 − nx + n(n−1)x^{2}/2! )
= 1 + (m−n)x + [(n^{2} − n)/2 − mn + (m2 − m)/2] x^{2}
Given m − n = 3 or n = m − 3
Hence (n^{2} − n)/2 − mn + (m^{2 }− m)/2 = 6
⇒ [(m−3)(m−4)]/2 − m(m−3) + (m^{2} − m)/2 = −6
⇒ m2 − 7m + 12 − 2m2 + 6m + m^{2} − m + 12 = 0
⇒ − 2m + 24 = 0
⇒ m = 12.
Find the largest coefficient in the expansion of (1 + x)^{n}, given that the sum of coefficients of the terms in its expansion is 4096. [REE 2000 (Mains)]
We know that, the coefficients in a binomial expansion is obtained by replacing each variable by unit in the given expression.
Therefore, sum of the coefficients in (a+b)^n
= 4096=(1+1)^{n}
⇒ 4096=(2)^{n}
⇒ (2)^{12}=(2)^{n}
⇒ n=12
Here n is even, so the greatest coefficient is nCn/2 i.e., 12C6
In the binomial expansion of (a  b)^{n}, n ³ 5, the sum of the 5th and 6th terms is zero. Then equals.
It is given that T6+T5=0.
Hence nC4a^{n}−4b^{4}−nC5a^{n}−5b^{5}=0
nC4a^{n}−4b^{4}= nC5a^{n}−5b^{5}
nC4a = nC5b
n!b/(n−4)!4! = n!b/(n−5)!5!
= a/(n−4) = b/5
Therefore a/b=(n−4)/5
Find the coefficient of x^{49} in the polynomial [REE 2001 (Mains), 3] where C_{r} = ^{50}C_{r}
The sum , (where = O if P < q) is maximum when m is [JEE 2002 (Scr.), 3]
(a) Coefficient of t^{24} in the expansion of (1 + t^{2})^{12} (1 + t^{12}) (1 + t^{24}) is [JEE 2003 (Scr.), 3]
(b) Prove that :
[JEE 2003 (Mains),2]
(1+x)^{12} (1+x^{12}) (1+x^{24})
= [C_{0} + C_{1}x^{2} + C_{2}x^{4} + C_{3}x^{6} + C_{4}x^{8} +.......C_{12}x^{24}) (1 + x^{12} + x^{24} + x^{36})
= x^{24}(^{12}C_{0} + ^{12}C_{6} + ^{12}C_{12})
= 1 + ^{12}C_{6} + 1
= ^{12}C_{6} + 2
^{n_1}C_{r} = (k^{2} – 3). ^{n}C_{r + 1}^{, }if k Î [JEE 2004 (Scr.)]
Formula,
(n  1)!/{r! × (n  1  r)!} = (k²  3) × n!/(r + 1)!(n  r  1)!
or, (n  1)!/r! = (k²  3) × n!/(r + 1)!
or, (n  1)!/r! = (k²  3) × n(n  1)!/(r + 1)r!
or, 1/1 = (k²  3) × n/(r + 1)
or, (r + 1)/n = (k²  3)
we know, r and n are integers so, (r + 1)/n (0, 1]
So, 0 < (r + 1)/n ≤ 1
or, 0 < k²  3 ≤ 1
or, 3 < k² ≤ 4
or, √3 < k ≤ 2 , 2 ≤ k < √3
Hence, k (√3, 2]
^{n1}C_{r} = (k^{2}  3). ^{n}C_{r + 1} [JEE 2004 (Scr.)]
(n  1)!/{r! × (n  1  r)!} = (k²  3) × n!/(r + 1)!(n  r  1)!
or, (n  1)!/r! = (k²  3) × n!/(r + 1)!
or, (n  1)!/r! = (k²  3) × n(n  1)!/(r + 1)r!
or, 1/1 = (k²  3) × n/(r + 1)
or, (r + 1)/n = (k²  3)
we know, r and n are integers so, (r + 1)/n (0, 1]
so, 0 < (r + 1)/n ≤ 1
or, 0 < k²  3 ≤ 1
or, 3 < k² ≤ 4
or, √3 < k ≤ 2 , 2 ≤ k < √3
hence, k (√3, 2]
The value of
is, where = ^{n}C_{r } [JEE 2005 (Scr.)]
The number of seven digit integers, with sum of the digits equal to 10 and formed by using the digits 1, 2 and 3 only, is [JEE 2009]
For r = 0, 1, ...., 10 let A_{r}, B_{r}, C_{r} denote, respectively, the coefficient of x^{r} in the expansions of (1 + x)^{10}, (1 + x)^{20} and (1 + x)^{30}. Then is equal to [JEE 2010]
Let a_{n} denote the number of all ndigit positive integers formed by the digits 0, 1 or both such that no consecutive digits in them are 0. Let b_{n} = the number of such ndigit integers ending with digit 1 and c_{n} = the number of such ndigit integers ending with digit 0. [JEE 2012]
Which of the following is correct ?
Let a_{n} denote the number of all ndigit positive integers formed by the digits 0, 1 or both such that no consecutive digits in them are 0. Let b_{n} = the number of such ndigit integers ending with digit 1 and c_{n} = the number of such ndigit integers ending with digit 0. [JEE 2012]
The value of b_{6} is
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