Simplify 02(n0)+12(n1)+22(n2)+⋯+n2(nn),
where n≥2.
Enter your answer in the form f(n)2g(n), where f(n) and g(n) are polynomials in n with integer coefficients.
Simplify
02(n0)+12(n1)+22(n2)+23(n2)+⋯+(n−1)2(nn−1)+n2(nn),
where n≥2.
S=02(n0)+12(n1)+22(n2)+23(n2)+⋯+(n−1)2(nn−1)+n2(nn)S=n∑k=0k2(nk)S=n∑k=0k∗k∗(nk)|k=(k1)S=n∑k=0k(k1)(nk)|(k1)(nk)=(n1)(n−1k−1)S=n∑k=1k(n1)(n−1k−1)|(n1)=nS=n∑k=1kn(n−1k−1)S=nn∑k=1k(n−1k−1)
n∑k=1k(n−1k−1)= ?
(1+x)n−1=(n−10)x0+(n−11)x1+(n−12)x2+⋯+(n−1n−1)xn−1|∗xx(1+x)n−1=(n−10)x1+(n−11)x2+(n−12)x3+⋯+(n−1n−1)xnx(1+x)n−1=n−1∑k=0(n−1k)xk+1ddxx(1+x)n−1=n−1∑k=0(k+1)(n−1k)xkx(n−1)(1+x)n−2+(1+x)n−1=n−1∑k=0(k+1)(n−1k)xk|x=1(n−1)(1+1)n−2+(1+1)n−1=n−1∑k=0(k+1)(n−1k)1k(n−1)(2)n−2+(2)n−1=n−1∑k=0(k+1)(n−1k)(n−1)2n−2+2n−1=n∑k=1k(n−1k−1)
S= ?
S=nn∑k=1k(n−1k−1)|n∑k=1k(n−1k−1)=(n−1)2n−2+2n−1S=n((n−1)2n−2+2n−1)S=n((n−1)2n−2+2n−1∗22)S=n((n−1)2n−2+2n−2∗2)S=n2n−2(n−1+2)S=n2n−2(n+1)S=n(n+1)2n−2|f(n)=n(n+1), g(n)=n−2, n≥2