AP EAMCET Engineering 21 Sep 2020 Shift 2 Paper

(c)

n

∑

r=1

r2Cr=12C1+22C2+32C3+......+n2Cn
∵(1+x)n=C0+C1x+C2x2+C3x3+.....+Cnxn
On differentiating both sides w.r.t. x, we get
n(1+x)n−1=1.C1+2C2x+3C3x2+.......+nCnxn−1
Now, on multiplying by x both sides, we get
nx(1+x)n−1=1.C1x+2C2x2+3C3x3+......+nCnxn
Now again on differentiating both sides w.r.t x, we get
n[(1‌+x)n−1+(n−1)x(1+x)n−2]
‌‌=12C1+22C2x+32C3x2+........+n2Cnxn−1
Put x=1, we get
12⋅C1+22C2‌+32C3+......+n2Cn
‌‌=n[2n−1+(n−1)2n−2]
‌‌=n2n−2[2+n−1]=n(n+1)2n−2