Ask your own question, for FREE!
Mathematics 7 Online
OpenStudy (anonymous):

Quick question about series... How would you go from: ∑k=0inf2(k+2)(k+1)ck+2xk+∑k=1infkckxk+∑k=0infckxk to \[\4c _{2} + c _{0} + sum_{k=1}^{inf}[2(k+2)(k+1)c _{k+2}+kc _{k}+c _{k}]x ^{k}\] I know we factored out the constant but 4c2 + c0 doesn't make too much sense to me. any help?

OpenStudy (amistre64):

∑(k=0..inf) [2(k+2)(k+1)c_k +2x^k] + ∑(k=1..inf) [k c_k x^k] + ∑(k=0..inf) [c_k x^k] \[\sum_{0}[2(k+2)(k+1)c_k+2x^k]+\sum_{1}[k~c_kx^k]+\sum_{0}[c_kx^k]\] is this what you got?

OpenStudy (amistre64):

if we want to get the index under the summations lined up, we need to pull out the 0th term so that they all have a starting index of 1 \[\sum_{0}[2(k+2)(k+1)c_k+2x^k]\\ ~~~~~~~~~~~~~~=2(0+2)(0+1)c_0+2x^0+\sum_{1}[2(k+2)(k+1)c_k+2x^k]\\ ~~~~~~~~~~~~~~=2(2)(1)c_0+2+\sum_{1}[2(k+2)(k+1)c_k+2x^k]\] \[\sum_{0}[c_kx^k]\\ ~~~~~~~~~~~~~~=c_0x^0+\sum_{1}[c_kx^k]\\ ~~~~~~~~~~~~~~=c_0+\sum_{1}[c_kx^k]\\ \] but its hard to read what you posted but this is the general process that i think they applied.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!