JEE maths #7

Algebra Level 4

Let C 0 , C 1 , C 2 . . . , C n C_0, C_1, C_2 ..., C_n be the binomial coefficients of the expansion ( 1 + x ) n (1+x)^n , where n ∈ N n \in \mathbb N . If S n = 2 C 0 2 + 2 2 C 1 6 + 2 3 C 2 12 + ⋯ + 2 n + 1 C n ( n + 1 ) ( n + 2 ) + 1 n + 1 + 1 2 ( n + 1 ) ( n + 2 ) S_n = \dfrac {2C_0}2 + \dfrac {2^2C_1}6 + \dfrac {2^3C_2}{12} + \cdots + \dfrac {2^{n+1}C_n}{(n+1)(n+2)} + \dfrac 1{n+1} + \dfrac 1{2(n+1)(n+2)} , then find the value of S 7 S 6 \dfrac {S_7}{S_6} .

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7 2 \frac 72 7 6 \frac 76 7 3 \frac 73 7 12 \frac 7{12} 7 7

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Rahil Sehgal by Rahil Sehgal , 20, India

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1 solution

Chew-Seong Cheong
Mar 30, 2017

S n = 2 C 0 2 + 2 2 C 1 6 + 2 3 C 2 12 + ⋯ + 2 n + 1 C n ( n + 1 ) ( n + 2 ) + 1 n + 1 + 1 2 ( n + 1 ) ( n + 2 ) = ∑ k = 0 n 2 k + 1 C k ( k + 1 ) ( k + 2 ) + 2 n + 5 2 ( n + 1 ) ( n + 2 ) = ∑ k = 0 n 2 k + 1 n ! ( k + 1 ) ( k + 2 ) k ! ( n − k ) ! + 2 n + 5 2 ( n + 1 ) ( n + 2 ) = ∑ k = 0 n 2 k + 1 n ! ( k + 2 ) ! ( n − k ) ! + 2 n + 5 2 ( n + 1 ) ( n + 2 ) = 1 2 ( n + 1 ) ( n + 2 ) ∑ k = 0 n 2 k + 2 ( n + 2 ) ! ( k + 2 ) ! ( n − k ) ! + 2 n + 5 2 ( n + 1 ) ( n + 2 ) = 1 2 ( n + 1 ) ( n + 2 ) ( ∑ k = 0 n ( n + 2 k + 2 ) 2 k + 2 + 2 n + 5 ) = 1 2 ( n + 1 ) ( n + 2 ) ( ∑ j = 0 n + 2 ( n + 2 j ) 2 j − ( n + 2 0 ) 2 0 − ( n + 2 1 ) 2 1 + 2 n + 5 ) = ( 1 + 2 ) n + 2 − 1 − 2 ( n + 2 ) + 2 n + 5 2 ( n + 1 ) ( n + 2 ) = 3 n + 2 2 ( n + 1 ) ( n + 2 ) \begin{aligned} S_n & = \frac {2C_0}2 + \frac {2^2C_1}6 + \frac {2^3C_2}{12} + \cdots + \frac {2^{n+1}C_n}{(n+1)(n+2)} + \frac 1{n+1} + \frac 1{2(n+1)(n+2)} \\ & = \sum_{k=0}^n \frac {2^{k+1}C_k}{(k+1)(k+2)} + \frac {2n+5}{2(n+1)(n+2)} \\ & = \sum_{k=0}^n \frac {2^{k+1}n!}{(k+1)(k+2)k!(n-k)!} + \frac {2n+5}{2(n+1)(n+2)} \\ & = \sum_{k=0}^n \frac {2^{k+1}n!}{(k+2)!(n-k)!} + \frac {2n+5}{2(n+1)(n+2)} \\ & = \frac 1{2(n+1)(n+2)} \sum_{k=0}^n \frac {2^{k+2}(n+2)!}{(k+2)!(n-k)!} + \frac {2n+5}{2(n+1)(n+2)} \\ & = \frac 1{2(n+1)(n+2)} \left(\sum_{k=0}^n {n+2 \choose k+2} 2^{k+2} + 2n+5 \right) \\ & = \frac 1{2(n+1)(n+2)} \left(\sum_{\color{#3D99F6}j=0}^{\color{#3D99F6}n+2} {n+2 \choose \color{#3D99F6}j} 2^{\color{#3D99F6}j} {\color{#3D99F6}- {n+2 \choose 0}2^0 - {n+2 \choose 1}2^1} + 2n+5 \right) \\ & = \frac {{\color{#3D99F6}(1+2)^{n+2} - 1 - 2(n+2)}+2n+5}{2(n+1)(n+2)} \\ & = \frac {3^{n+2}}{2(n+1)(n+2)} \end{aligned}

S 7 S 6 = 3 9 2 ( 8 ) ( 9 ) × 2 ( 7 ) ( 8 ) 3 8 = 7 3 \begin{aligned} \frac {S_7}{S_6} & = \frac {3^9}{2(8)(9)} \times \frac {2(7)(8)}{3^8} = \boxed{\dfrac 73} \end{aligned}

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Absolutely brilliant solution sir !!!

Rahil Sehgal - 4 years, 2 months ago

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You should learn up LaTex since you are posting so many problems. I have been converting your image problems into LaTex. You can see the codes by clicking the pull-down menu ⋯ \cdots at the right-hand bottom corner of the problem and select Toggle LaTex . Or place your mouse cursor on the formulas.

Chew-Seong Cheong - 4 years, 2 months ago

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Thank you very much sir... I have uploaded a new problem in LATEX. I hope the problem is much better...

Rahil Sehgal - 4 years, 2 months ago

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