The correct option is C 7
10∑r=2 rC2⋅ 10Cr=10∑r=2r(r−1)2× 10Cr
=10∑r=2r2−r2× 10Cr
=10∑r=212×r2× 10Cr−1210∑r=2r× 10Cr
(1+x)n= nC0+ nC1x+ nC2x2+...+ nCnxn
Differentiating both sides, we get
n(1+x)n−1= nC1+ 2⋅ nC2x+3⋅ nC3x2+...+n⋅ nCnxn−1 ...(1)
Let x=1
n×2n−1= nC1+2⋅ nC2+3⋅ nC3+4⋅ nC4+...+n⋅ nCn
=n∑r=1r nCr
Eqn. (1)×x
n x(1+x)n−1=x⋅ nC1+2x2⋅ nC2+3x3⋅ nC3+...+n xn⋅ nCn
Differentiating
n(1+x)n−1+n(n−1)x(1+x)n−2= nC1+22 nC2 x+32 nC3 x2+...+n2 xn−1 nCn
Put x=1
n⋅2n−1+n(n−1)2n−2=12 nC1+22 nC2+...+n2 nCn
⇒n⋅2n−1+n(n−1)2n−2=n∑r=1r2 nCr
∴ 10∑r=212r2 10Cr−1210∑r=2r⋅ 10Cr
=12[10⋅29+10⋅9⋅28−10]−12[10⋅29−10]
=10⋅28+10⋅9⋅27−10⋅28
=10⋅9⋅27
which is not divisible by 7.