Let Sn=n∑k=1k⋅2n+1−k=2n+1n∑k=1k⋅2−k
Let tn=n∑k=1k2k
⇒tn=12+222+323+⋯+n2n −−(1)
⇒tn2= 122+223+⋯+n2n+1 −−(2)
From (1)−(2), we get
tn2=12+122+123+⋯+12n−n2n+1
⇒tn2=12(1−(12)n)1−12−n2n+1
⇒tn=2(1−12n−n2n+1)
∴Sn=2n+1×tn
=2n+1×2(1−12n−n2n+1)
=2[2n+1−2−n]
On comparing with the expression (n+14)(2n+1−n−2),
n+14=2⇒n=7