Show that for k running over positive integers ∑k=1∞k22k=6

Question

Show that for k running over positive integers  ∑k=1∞k22k=6

nick1337 · Accepted Answer

Alternative way of solving this problem. Using some basic calculus of the power series. Starting from ∑k≥1k2(12)k, consider a power series ∑k≥1k2xk=x∑k≥1k2xk−1 which converges uniformly whenever |x|&amp;lt;1. We get our sum by standard differentiation and integration of the power series. You can integrate ∑k≥1k2xk−1 term by term so ∑k≥1∫0xk2tk−1dt=∑k≥0kxk. Again, ∑k≥0kxk=1+x∑k≥1kxk−1, by integrating term by term ∑k≥1∫0xktk−1dt=∑k≥0xk=11−x. Now, we must take the derivative so (11−x)′=1(1−x)2 so1+x∑k≥1kxk−1=1+x(1−x)2Taking again the derivative (1+x(1−x)2)′=1+x(1−x)3 so∑k≥1k2xk=x(1+x)(1−x)3Finally, by putting x=12 we get that ∑k≥1k2(12)k=6

star233 · Accepted Answer

Consider the series ∑k=0∞xk2k=11−x/2 Take a derivative ∑k=0∞kxk−12k=1/2(1−x/2)2 (1) Take another derivative ∑k=0∞k(k−1)xk−22k=1/2(1−x/2)3 (2) Adding (1) and (2) at x=1, we get ∑k=0∞k22k=6

RizerMix · Accepted Answer

For convenience, we denote ak=k2+2k+32k−1, k=1,2,... Then k22k=ak−ak+1 Thus ∑k=1nk22k=(a1−a2)+(a2−a3)+...+(an−an+1)=a1 =6−(n+1)2+2(n+1)+32n =6−n2+4n+62n It follows that ∑k=1∞k22k=limn→∞∑k=1nk22k =limn→∞(6−n2+4n+62n =6−0 =6

Show that for k running over positive integers \sum_{k=1}^\infty\frac{k^2}{2^k}=6

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