0
$\begingroup$

Can someone explain simply why $\displaystyle\sum_{k = 1}^n\dfrac{k^2}{2^{k - 1}} = \displaystyle\sum_{k = 0}^{n - 1}\dfrac{\left(k + 1\right)^2}{2^k}$? I don't get why we go from $k$ to $k+1$ in the numerator, please help me understand it.

$\endgroup$
5
$\begingroup$

It looks like you are asking why $$\sum_{k=1}^n \frac{k^2}{2^{k-1}}=\sum_{k=0}^{n-1} \frac{(k+1)^2}{2^k}.$$ This is a change of summation index. We have $1\le k \le n$, so $0\le k-1 \le n-1$. Now let $j=k-1$, so $0\le j \le n-1$, and $$\sum_{k=1}^n \frac{k^2}{2^{k-1}}=\sum_{k-1=0}^{n-1} \frac{(k-1+1)^2}{2^{k-1}}=\sum_{j=0}^{n-1} \frac{(j+1)^2}{2^j}.$$ Now just change $j$ to $k$ to obtain $$\sum_{k=0}^{n-1} \frac{(k+1)^2}{2^k}.$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ This is very helpful! Thank you Rob $\endgroup$ – Henry Huynh Jan 26 at 17:50
  • $\begingroup$ Glad to help. Please mark the answer accepted. $\endgroup$ – RobPratt Jan 26 at 19:21
  • $\begingroup$ Done! Thanks again $\endgroup$ – Henry Huynh Jan 26 at 20:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.