Evaluate ∑k=1∞k2(k−1)!

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sphwngzt · Accepted Answer

Let&#039;s start ex=∑k=1∞xk−1(k−1)! then:xex=∑k=1∞xk(k−1)!x(xex)′=xex+x2ex=∑k=1∞kxk(k−1)!x(x(xex)′)′=x(ex+3xex+x2ex)=∑k=1∞k2xk(k−1)!Set x=1 to get:5e=∑k=1∞k2(k−1)!

jojoann325 · Accepted Answer

The most elementary calculation is probably this one:  ∑k≥0(k+1)2k!=1+∑k≥1k2+2k+1k!  =1+∑k≥1k(k−1)!+2∑k≥11(k−1)!+∑k≥11k!  =∑k≥1k−1+1(k−1)!+2∑k≥01k!+∑k≥01k!  =1+∑k≥21(k−2)!+∑k≥21(k−1)!+3∑k≥01k!  =∑k≥01k!+4∑k≥01k!  =5∑k≥01k!  =5e.  One can also make use of the identity  xn=∑k(3i)ki  =0⋅k0+1⋅k1+3⋅k2+1⋅k3  =k+3k(k−1)+k(k−1)(k−2)  Whence  k2(k−1)!=k3k!=k+3k(k−1)+k(k−1)(k−2)k!  =1(k−1)!+3(k−2)!+1(k−3)!,  and summing over l yields 5e as before.

RizerMix · Accepted Answer

For P(n)(n−r)! where P(n) is a polynomial. If the degree of P(n) is m&amp;gt;0, we can write P(n)=A0+A1(n−r)+A2(n−r)(n−r−1)+...+Am(n−r)(n−r−1)...{(n−r)−(m−1)} Here k2=C+B(k−1)+A(k−1)(k−2) Putting k=1 in the above identity, C=1 k=2, B+C=4⇒ B=3 k=0⇒2A−B+C=0,2A=B−C=3−1⇒A=1 or comparing the coefficients of k2, A=1 So, k2(k−1)!=(k−1)(k−2)+3(k−1)+1(k−1)!=1(k−3)!+3(k−2)!+1(k−1)! ∑k=1∞k2(k−1)!=∑k=1∞(1(k−3)!+3(k−2)!+1(k−1)! ∑k=1∞k2(k−1)!=∑k=1∞((k−1)(k−2)+3(k−1)+1(k−1)!) =∑k=3∞1(k−3)!+∑k=2∞3(k−2)!+∑k=1∞1(k−1)! as 1(−r)!=0 for r&amp;gt;0 =e+3e+e=5e

Evaluate ∑k=1∞k2(k−1)!

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