Question

If $C_0,C_1,C_2,\dots ,C_n$ denotos the binomial coefficients in the expansion of $(1+x{)}^n,$ and $1^3\cdot C_1+2^3\cdot C_2+3^3\cdot C_3+\dots +n^3{C}_n=\frac{1}{\lambda }\left(n^2\right)(n+\mu )\cdot 2^n,$ then $\lambda +\mu =$

Answer 1

Let
$S_n=1^3\cdot C_1+2^3\cdot C_2+3^3\cdot C_3+\dots +n^3{C}_n$

$=\sum _{r=1}^n{r}^3\cdot C_r$

$=\sum _{r=1}^n\left(r\right(r-1\left)\right(r-2)+3r(r-1)+r)\cdot C_r$

$=\sum _{r=1}^{n}r(r-1)(r-2){}^n{C}_r+3\sum _{r=1}^{n}r(r-1){}^n{C}_r+\sum _{r=1}^{n}r\cdot {}^n{C}_r$

$=\sum _{r=3}^{n}n(n-1)(n-2){}^{n-3}{C}_{r-3}+3\sum _{r=2}^{n}n(n-1){}^{n-2}{C}_{r-2}+\sum _{r=1}^{n}n\cdot {}^{n-1}{C}_{r-1}$

$[\because \frac{{}^n{C}_r}{{}^{n-3}{C}_{r-3}}=\frac{n(n-1)(n-2)}{r(r-1)(r-2)},\frac{{}^n{C}_r}{{}^{n-2}{C}_{r-2}}=\frac{n(n-1)}{r(r-1)},\frac{{}^n{C}_r}{{}^{n-1}{C}_{r-1}}=\frac{n}{r}]$

$=n(n-1)(n-2)\sum _{r=3}^n{}^{n-3}{C}_{r-3}+3n(n-1)\sum _{r=2}^n{}^{n-2}{C}_{r-2}+n\sum _{r=1}^n{}^{n-1}{C}_{r-1}$

$=n(n-1)(n-2)\cdot 2^{n-3}+3n(n-1)\cdot 2^{n-2}+n\cdot 2^{n-1}$

$=n\cdot 2^{n-3}\left[\right(n-1\left)\right(n-2)+6(n-1)+4]$

$=n\cdot 2^{n-3}[n^2+3n]$

$=n^2\cdot 2^{n-3}(n+3)$

$=\frac{1}{8}\cdot n^2(n+3)\cdot 2^n$

So, $\lambda =8,\mu =3$

$\Rightarrow \lambda +\mu =8+3=11$