Prove that: ∑n=1∞n2(n−1)2n=20

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Prove that:  ∑n=1∞n2(n−1)2n=20

Troy Sutton · Accepted Answer

Note that ∑ntn=11−t for every |t|&amp;lt;1 hence, differentiaiting this twice and three times,  ∑nn(n−1)tn−2=2(1−t)3,  ∑nn(n−1)(n−2)tn−3=6(1−t)4  For t=12, this reads  ∑nn(n−1)12n−2=2⋅23  ∑nn(n−1)(n−2)12n−3=6⋅24  which implies  ∑nn(n−1)12n=14⋅2⋅23=4  ∑nn(n−1)(n−2)12n=18⋅6⋅24=12  Finally,  n2(n−1)=2⋅n(n−1)+n(n−1)(n−2)  hence  ∑nn2(n−1)12n=2⋅∑nn(n−1)12n+∑nn(n−1)(n−2)12n=2⋅4+12  This approach can be made shorter if one notices once and for all that, for every nonnegative k,  ∑nn(n−1)…(n−k)12n=2⋅(k+1)!

egowaffle26ic · Accepted Answer

For |x|&amp;lt;1 we have that ∑n=0∞xn=11−x hence differentiating both sides we get  ∑n=0∞(n+1)xn=∑n=1∞nxn−1=1(1−x)2  while differentiating once more  ∑n=0∞(n+2)(n+1)xn=2(1−x)3  and once again  ∑n=0∞(n+3)(n+2)(n+1)xn=6(1−x)4  Next we express n2(n−1) as a linear combination of (n+3)(n+2)(n+1),(n+2)(n+1),(n+1) and 1:  n2(n−1)=(n+3)(n+2)(n+1)−7(n+2)(n+1)+10(n+1)−2  and hence  ∑n=0∞n2(2n−1)xn=6(1−x)4−14(1−x)3+10(1−x)2−21−x  and setting x=12 we obtain that  ∑n=0∞n2(n−1)2n=6⋅16−4⋅8+10⋅4−4=20

Prove that: \sum_{n=1}^\infty\frac{n^2(n-1)}{2^n}=20

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