Question

If $S_n=^nC_0.^nC_1+^nC_1.^nC_2+....+^nC_{n-1}.^nC_n\;and\;\dfrac{S_{n+1}}{S_n}=\dfrac{15}4,\;$ then n is
A.8
B.6
C.4
D.16

Answer 1

Hint : One should have the knowledge of binomial theorem to solve this problem. Some basic formula are given below , we make use of this formulae to solve the question.
\begin{aligned} & ^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}...\left( 1 \right) \\\ &{\left( {1 + x} \right)^n}{ = ^n}{C_0}{.1^n}.{x^0}{ + ^n}{C_1}{.1^{n - 1}}.{x^1} + {....^n}{C_n}{.1^0}.{x^n}\;\;\;\;...\left( 2 \right) \\\ &= \mathop \sum \limits_{r = 0}^n {}_{}^nC_r^{}{.1^{n - r}}.{x^r} \\\ \end{aligned}

Complete step-by-step answer :
The given summation $S_n$ is a sum of products of the terms of two different series. So, we will find these series and express $S_n$ in terms of that series-

By equation (2), we can write that-
$\left(1+x\right)^n=^nC_0.1^n.x^0+^nC_1.1^{n-1}.x^1+....^nC_n.1^0.x^n\\\\\left(1+x\right)^n=^nC_0+^nC_1.x+^nC_2.x^2....^nC_n.x^n\\\Also,\;when\;we\;replace\;x\;by\;\dfrac1x\\\\\left(1+\dfrac1x\right)^n=^nC_0.1^n.\dfrac1{x^0}+^nC_1.1^{n-1}.\dfrac1{x^1}+....^nC_n.1^0.\dfrac1{x^n}\\\\\left(1+\dfrac1x\right)^n=^nC_0+^nC_1.\dfrac1x+^nC_2.\dfrac1{x^2}....^nC_n.\dfrac1{x^n}$

On multiplying these two binomial series, we can write that-
\left(1+\mathrm x\right)^{\mathrm n}\left(1+\dfrac1{\mathrm x}\right)^{\mathrm n}=\left({}^{\mathrm n}{\mathrm C}_0+^{\mathrm n}{\mathrm C}_1.\mathrm x+^{\mathrm n}{\mathrm C}_2.\mathrm x^2....^{\mathrm n}{\mathrm C}_{\mathrm n}.\mathrm x^{\mathrm n}\right)\left({}^{\mathrm n}{\mathrm C}_0+^{\mathrm n}{\mathrm C}_1.\dfrac1{\mathrm x}+^{\mathrm n}{\mathrm C}_2.\dfrac1{\mathrm x^2}....^{\mathrm n}{\mathrm C}_{\mathrm n}.\dfrac1{\mathrm x^{\mathrm n}}\right)\\\
When we look at the series closely, the coefficients of x in the product are the same as $S_n$ . The second term in the first series multiplies with the first term in the second series to give x. Similarly, the third term in the first series multiplies with the second term in the second series and so on. If we look at the coefficients, they are the same as $S_n$ .

$S_n$ = coefficient of x in $\left(1+x\right)^n\left(1+\dfrac1x\right)^n$

$S_n$ = coefficient of x in $\dfrac{\left(1+\mathrm x\right)^{2\mathrm n}}{\mathrm x^{\mathrm n}}\left(\mathrm{By}\;\mathrm{multiplying}\;\mathrm{within}\;\mathrm{the}\;\mathrm{brackets}\right)$

$S_n$ = coefficient of $x^{n+1}$ in $(1 + x)^{2n}$ (By multiplying both sides by $x^n$ )

The formula for $n^{th}$ term in binomial expression is given by-
{}^{\mathrm n}{\mathrm C}_{\mathrm r}.1^{\mathrm n-\mathrm r}\mathrm x^{\mathrm r}\\\\\mathrm{So},\;\\\\\mathrm{Coefficient}\;\mathrm{of}\;\mathrm x^{\mathrm n+1}=^{2\mathrm n}{\mathrm C}_{\mathrm n+1}.1^{2\mathrm n-\left(\mathrm n+1\right)}\mathrm x^{\mathrm n+1}\\\=^{2\mathrm n}{\mathrm C}_{\mathrm n+1}
So, we can write that-
{\mathrm S}_{\mathrm n}=^{2\mathrm n}{\mathrm C}_{\mathrm n+1}=\dfrac{\left(2\mathrm n\right)!}{\left(\mathrm n-1\right)!\left(\mathrm n+1\right)!}\\\\{\mathrm S}_{\mathrm n+1}=^{2\mathrm n+2}{\mathrm C}_{\mathrm n+2}=\dfrac{\left(2\mathrm n+2\right)!}{\left(\mathrm n\right)!\left(\mathrm n+2\right)!}\\\
We have been given the ratio of $S_{n+1}$ and $S_n$ , so we can equate that with these values to find the value of n.
$\dfrac{{{S_{n + 1}}}}{{{S_n}}} = \dfrac{{15}}{4} = \dfrac{{\left( {\dfrac{{(2n + 2)!}}{{n!(n + 2)!}}} \right)}}{{\left( {\dfrac{{(2n)!}}{{(n - 1)!(n + 1)!}}} \right)}}$
$\dfrac{{15}}{4} = \dfrac{{\left( {\dfrac{{1.2.3...(2n)(2n + 1)(2n + 2)}}{{[1.2.3...(n - 1)n][1.2.3...(n + 1)(n + 2)]}}} \right)}}{{\left( {\dfrac{{1.2.3...(2n)}}{{[1.2.3...(n - 1)][1.2.3...(n + 1)]}}} \right)}}$
We can cancel out the factorials correspondingly as-
$\dfrac{{(2n + 1)(2n + 2)}}{{n(n + 2)}} = \dfrac{{15}}{4}$
$4(4{n^2} + 4n + 2n + 2) = 15({n^2} + 2n)$
$16{n^2} + 24n + 8 = 15{n^2} + 30n$
${n^2} - 6n + 8 = 0$
${n^2} - 4n - 2n + 8 = 0$
n(n - 4) - 2(n - 4) = 0
(n - 4)(n - 2) = 0
n = 2, 4
2 is not in the options, so the correct answer is C. 4

Note : These problems require a deep understanding of binomial expressions and expansion. This is because we have to analyze the expression to write $S_n$ in terms of n. After we do this, we can solve the problem easily with simple calculations.