Find: $ \sum\limits_{k=1}^n \left(k^3 - (k-1)^3\right)$

Question

Find: $ \sum\limits_{k=1}^n \left(k^3 - (k-1)^3\right)$

I tried this problem using the definition of the sum but I always end up getting the wrong answer. Im trying not to use induction in this problem but straight logic to find out. Any help would be appreciated.

Answer 1

${\bf Hint.}$ This is telescoping. See your sum equals

$$ (1^3 - 0^3) + (2^3-1^3) + (3^3 - 2^3) + ... + ((n-1)^3 - (n-2)^3)+ (n^3 - (n-1)^3) = $$

In particular $\sum_{k=1}^n (a_k - a_{k-1}) = a_n - a_0 $. So, in your case, ans = $\boxed{ n^3}$

Answer 2

Two possible things to recognize:

1) One

$k^3 - (k-1)^3 = k^3 - (k^3 -3k^2 + 3k -1) = 3k^2-3k + 1$ and we know $\sum_{k=1}^n k^2 = \frac {n(n+1)(2n+1)}6$ and $\sum_{k=1}^n k = \frac {n(n+1)}2 $ and $\sum_{k=1}^n 1 = n$.

So then is $\frac {n(n+1)(2n+1)}2 - \frac {3n(n+1)}2 + n$.

$= \frac {n(n+1)(2n+1 - 3) + 2n}2 = \frac {n(n+1)(2n-2) + n}2 = n(n+1)(n-1) + n = n(n^2 -1) + n = n^3 - n + n = n^3$

2) Two

If $a_k = k^3$ this $\sum_{k=1}^n (a_k - a_{k-1})$ which is a telescoping series:

$\sum_{k=1}^n (a_k - a_{k-1}) = \sum\limits_{j=k=1}^n a_j - \sum\limits_{j=k-1=0}^{n-1} a_j = a_n - a_0 = n^3 -0^2 = n^3$.

Answer 3

The values inside the brackets simplify to $3k^2-3k+1$. Consider a combinatoric approach where $6{k \choose 2}=3(k)^2 -3k$.

Thus we replace these getting $6\displaystyle\sum_{k=1}^{n}{k \choose 2}+\displaystyle\sum_{k=1}^{n}1 $

=$6{n+1 \choose 3}+n$

=$6( \frac{n+1!}{3!(n-2)!})+n$

=$(n+1)(n)(n-1)-n$ = $n^3$