Question #d1c04

Question

1081 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-10T17:36:53

#2n(n+1)+1#

For #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#