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## Telescoping Sums and Products (Solutions 1)

Here is the solution provided for this previous problem on Telescoping Sums and Products

### Example 1.  Prove that

$\displaystyle\sum^{n}_{k=1}k^4=\dfrac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$

### Example 1.  Solution

From the Binomial Theorem we have

$\displaystyle(k+1)^5-k^5=5k^4+10k^3+10k^2+5k+1$

thus

$\displaystyle\sum^{n}_{k=1}[(k+1)^5-k^5]=(n+1)^5-1$

It follows that

$\displaystyle 5\sum^{n}_{k=1}k^4+10 \sum^{n}_{k=1}k^3+10 \sum^{n}_{k=1}k^2+5 \sum^{n}_{k=1}k+\sum^{n}_{k=1}1=(n+1)^5-1$

Hence

$\displaystyle5 \sum^{n}_{k=1}k^4=(n+1)^5-10\dfrac{n(n+1)(2n+1)}{6}-5\dfrac{n(n+1)}{6}-(n+1)$

Yielding

$\displaystyle\sum^{n}_{k=1}k^4=\dfrac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$