IIT JEE , AIEEE Forum - Mathematics , Algebra

ramkumar_november

the given expression is....

$\sum_{m=1}^n\Bigg(\sum_{k=1}^m\Bigg(\sum_{p=k}^m\;nC_m.mC_p.pC_k \Bigg)\Bigg)$

= $\sum_{m=1}^n nC_m\Bigg(\sum_{k=1}^m\Bigg(\sum_{p=k}^m\;\frac{m!}{p!(m-p)!}.\frac{p!}{k!(p-k)!} \Bigg)\Bigg)$

= $\sum_{m=1}^n nC_m\Bigg(\sum_{k=1}^m\Bigg(\sum_{p=k}^m\; ^{m-k}C_{p-k} \Bigg)\Bigg)\frac{m!}{k!(m-k)!}$

= $\sum_{m=1}^n nC_m\Bigg(\sum_{k=1}^m\;2^{m-k}. ^m C_k \Bigg)$

= $\sum_{m=1}^n nC_m\Bigg((1+2)^m-2^m\Bigg)$

= $\sum_{m=1}^n ^n C_m.3^m - ^n C_m.2^m$

=

=

so

$\sum_{m=1}^n\Bigg(\sum_{k=1}^m\Bigg(\sum_{p=k}^m\;nC_m.mC_p.pC_k \Bigg)\Bigg)$ = 4^n-3^n

please nudge me if you have doubts in any of the steps......