jee question!!!

we\; have\; to\; equate;\; \frac{n+1}{4}[2^{n+1}-n-2]=\sum_{0}^{n}{k(2^{n+1-k})}=2^{n+1}.\sum_{0}^{n}{\frac{k}{2^k}}

i.e.\;\frac{n+1}{4}[2^{n+1}-n-2]=2^{n+1}.\sum_{0}^{n}{\frac{k}{2^k}}

NOW, \sum_{0}^{n}{\frac{k}{2^k}}=2\frac{2^{n+1}-n-2}{2^{n+1}}

so, \frac{n+1}{4}[2^{n+1}-n-2]=2^{n+1}.2.\frac{2^{n+1}-n-2}{2^{n+1}}

comparing we get, \frac{n+1}{4}=2\;\;\; \;or,\;n=7

to prove Σ(k2^k) = 2ⁿ⁺¹-n-22ⁿ⁺¹

consider, S=Σ(k2^k)

S2=12.Σ(k2^k)=Σ(k2^k+1)

subtracting the 2 expressions we get, S2=12 + {14 + 18 + 116 + ... (n-1) terms} - n2ⁿ⁺¹

or, S2={12 + 14 + 18 + 116 + ... n terms} - n2ⁿ⁺¹

or, S2=12(1-2^-n)1-(1/2) - n2ⁿ⁺¹=(2ⁿ-12ⁿ) - n2ⁿ⁺¹=2.2ⁿ-n-22ⁿ⁺¹=2ⁿ⁺¹-n-22ⁿ⁺¹....... HENCE PROVED!!! [1][1][1]