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Given positive integers r > 1, n >2 and that the coefficient of (3r)^{th} and (r + 2)^{th} terms in the binomial expansion of (1+ x)^{2n} are equal . Then (1983  1 Mark)
Given that r and n are +ve integers such that r > 1, n > 2
Also in the expansion of (1+ x)^{2n} coeff. of (3r)^{th} term = coeff. of (r + 2)^{th }term
⇒ ^{2n}C_{3r–1}= ^{2n}C_{r+1}
⇒ 3r –1 = r + 1 or 3r – 1+ r + 1 = 2n
⇒ r = 1 or 2r = n
But r > 1
∴ n = 2r
The coefficient of x^{4} in is (1983  1 Mark)
General term in the expansion is
For coeff of x^{4}, we should have
10 – 3r = 4 ⇒ r = 2
∴ Coeff of x^{4} =
The expression is a polynomial of degree (1992  2 Marks)
The given expression is
We know by binomial theorem, that (x + a ) n + ( x – a )^{ n} = 2 [ ^{n}C_{0 }x^{ n} + ^{n}C_{ 2}x ^{n} ^{– 2}a ^{2 }+ ^{n}C_{4}x^{n} ^{– 4}a_{4} + ......]
∴ The given expression is equal to 2 [^{5}C_{0}x^{5} + ^{ 5}C_{2}x_{3}(x^{3} – 1) + ^{5}C_{4}x (x^{3} – 1)^{2}]
Max. power of x involved here is 7, also only +ve integral powers of x are involved, therefore given expression is a polynomial of degree 7.
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 (1999  2 Marks)
We have (1 + x)^{m} (1– x)^{n}
= 1 + (m + n)x +
Given, m – n = 3 ....(1)
and
⇒ m^{2} + n^{2}  2mn  (m +n) =12
⇒ (m n)^{2}  (m +n) =12
⇒ m +n = 9 + 12= 21 ....(2)
From (1) and (2), we get m = 12
For (2000S)
NOTE THIS STEP :
In the binomial expansion of (a  b)^{n}, n ≥ 5, the sum of the 5^{th} and 6^{th} terms is zero. Then a/b equals (2001S)
(a – b)^{n}, n ≥ 5 In binomial expansion of above T5+ T6= 0
⇒ ^{n}C_{4} a^{n–4} b_{4} + ^{n}C_{5 }a^{n–5} b^{5} = 0
The sum (where is maximum when m is (2002S)
= Coeff of x^{m} in the expansion of product (1+ x)^{10} (1 + x)^{20} = Coeff of x^{m} in the expansion of (1+ x)^{30} = ^{30}C_{m} To get max. value of given sum, 30Cm should be max. which is so when m = 30/2 = 15.
Coefficient of t^{24} in (1 +t^{2})^{12} (1+t^{12}) (1 + t^{24}) is (2003S)
(1 + t^{2})^{12} (1 + t^{12}) (1 + t^{24}) = (1 + t^{12} + t^{24} + t^{36}) (1 + t^{2})^{12 }
∴ Coeff. of t^{24 }= 1× Coeff. of t^{24} in (1+ t^{2})^{12 }+ 1 × Coeff. of t^{12} in (1 + t^{2})^{12 }+ 1 × constant term in (1 + t^{2})^{12} = ^{12}C_{12 }+ ^{12}C_{6} + ^{12}C_{0} = 1+^{12}C_{6} + 1=^{12}C_{6} + 2
If ^{n–1}C_{r }= (k^{2} – 3) ^{n}C_{r +1}, then k ∈ (2004S)
Since 0 ≤ r ≤ n – 1
The value of is where (2005S)
To find ^{30}C_{0}^{30}C_{10} – ^{30}C_{1}^{30} C_{11 }+ ^{30} C_{2}^{30}C_{12} – .... + ^{30}C_{20}^{30}C_{30}
We know that (1 + x)^{30 }= ^{30}C_{0} + ^{30}C_{1}x + ^{30}C_{2}x^{2} + .... + ^{30}C_{20}x^{20} + ....^{30}C_{30}x^{30} ....(1)
(x – 1)^{30} = ^{30}C_{0}x^{30} – ^{30}C_{1}x^{29 }+....+ ^{30}C_{10}x^{20} – ^{30}C_{11}x^{19} + ^{30}C_{12}x^{18 }+.... ^{30}C_{30}x^{0} ....(2)
Multiplying eqn (1) and (2), we get (x^{2} – 1)^{30} = ( ) × ( )
Equating the coefficients of x^{20} on both sides,
we get ^{30}C_{10} = ^{30}C_{0}^{30} C_{10} – ^{30} C_{1}^{30}C_{11 } + ^{30}C_{2}^{30}C_{12}– ....
+ ^{30}C_{20} ^{30}C_{30}
∴ Req. value is ^{30}C_{10}
For r = 0, 1, …, 10, let A_{r}, B_{r }and C_{r} denote, respectively, the coefficient of x^{r} in the expansions of (1 + x)^{10} , (2010)
(1 + x)^{20 }and (1 + x)^{30}. Then is equal to
Clearly
Now expanding (1 + x )^{10 }and (1 + x )^{20 }by bin om ial theorem and comparing the coefficients of x^{20 }in their product, on both sides, we get
= coeff of x^{20} in (1 + x )^{30} = ^{30} C_{20}= ^{30}C_{10}
Again expending (1 + x )^{10} and ( x + 1)^{10} by bin omial theorem and comparing the coefficients of x^{10} in their
product on both sides, we get
coeff of x^{10} in (1 + x)^{ 20 }= _{20}C_{10}
Substituting these values in equation (1), we get
Coefficient of x^{11} in the expansion of (1 + x^{2})^{4}(1 + x^{3})^{7 }(1 + x^{4})^{12} is (JEE Adv. 2014)
Coeff. of x^{11 }in exp. of
= (Coeff. of x^{a} ) × (Coeff. of x^{b} ) × (Coeff. of x^{c} )
Such that a + b + c = 11 Here a = 2m, b = 3n, c = 4p
∴ 2m + 3n + 4p = 11
Case I : m = 0, n = 1, p = 2
Case II : m = 1, n = 3, p = 0
Case III : m = 2, n = 1, p = 1
Case IV : m = 4, n = 1, p = 0
∴ Required coeff.
= 462 + 140 + 504 + 7 = 1113
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