Question #d1c04

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Answer 1 · 2017-04-10T17:36:53

$2 n (n + 1) + 1$ $2n(n+1)+1$

For $| x | < 1$ $abs x < 1$ we have

$lim_(n->oo)(1-x^(n+1))/(1-x) = lim_(n->oo)sum_(k=0)^n x^k = 1/(1-x)$

and also

$(d^2)/(dx^2)(1/(1-x))=2/(1-x)^3=sum_(k=2)^oo k(k-1)x^(k-2)$

so, for $abs x < 1$

$(1+x)^2/(1-x)^3 = 1/2(x^2+2x+1)sum_(k=2)^oo k(k-1)x^(k-2)=$

$=1/2(sum_(k=2)^oo k(k-1)x^k+2sum_(k=2)^oo k(k-1)x^(k-1)+sum_(k=2)^oo k(k-1)x^(k-2))$ and for $k ge 2$

$(1+x)^2/(1-x)^3=1/2sum_(k=2)^oo((k(k-1)+2k(k+1)+(k+1)(k+2))x^k$

or for $k ge 2$

$(1+x)^2/(1-x)^3=sum_(k)^oo(2k(k+1)+1)x^k$ so the coefficient for $x^n$

is

$2n(n+1)+1$