(a) Binomial expansion is:
(1+x)ⁿ=1+nx+n(n-1)x²/2!+n(n-1)(n-2)x³/3!+...+C(n,k)xᵏ
where C(n,k)=n!/[k!(n-k)!], so (1+x)ⁿ=1+∑[k=1,n]C(n,k)xᵏ.
The derivative of (1+x)ⁿ is n(1+x)ⁿ⁻¹, so:
n(1+x)ⁿ⁻¹=∑[k=1,n]kC(n,k)xᵏ⁻¹, given that C(n,r)=(ⁿᵣ).
(b) When x=1, n(1+1)ⁿ⁻¹=n×2ⁿ⁻¹=∑[k=1,n]kC(n,k) because 1ᵏ=1.
(c) Derivative of n(1+x)ⁿ⁻¹=n(n-1)(1+x)ⁿ⁻²=∑[k=1,n]k(k-1)C(n,k)xᵏ⁻².
When x=1, n(n-1)(2ⁿ⁻²)=∑[k=1,n]k(k-1)C(n,k).
n×2ⁿ⁻¹=2n×2ⁿ⁻²=2∑[k=1,n]kC(n,k)
k(k-1)+2k=k(k+1), so:
∑[k=1,n]k(k-1)C(n,k)+2∑[k=1,n]kC(n,k)=∑[k=1,n]k²C(n,k),
n(n-1)(2ⁿ⁻²)+2n×2ⁿ⁻²=∑[k=1,n]k²C(n,k),
n(n+1)2ⁿ⁻²=∑[k=1,n]k²C(n,k).