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## Telescoping Sums and Products (Example 2)

This time I will provide both the problem and solution in  the continuing series from Dr. Andreescu’s lecture on Telescoping Sums and Products.

### Example 2.  Evaluate $\displaystyle\sum^{n}_{k=2}k!(k^2+k+1)$

### Example 2.  Solution

We can write $\displaystyle\sum^{n}_{k=1}k!(k^2+k+1)=\displaystyle\sum^{n}_{k=1}[(k+1)^2-k)]k!$ $=\displaystyle\sum^{n}_{k=1}[(k+1)!(k+1)-k!k]$ $=(n+1)!(n+1)-1$

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

## Telescoping Sums and Products (Example 1)

While we wait for the Metroplex Math Circle to continue again in mid January, I will be posting a series of problems and solutions from our last session. Those who were fortunate enough to attend learned useful techniques for solving problems common in math contests.

### Telescoping Sums and Products

The telescoping sums and products idea is used to solve many problems involving sums or products in algebra. For problems involving sums, the idea is to use identities, to write the sum in the form $\displaystyle\sum^{n}_{k=1}[F(k+1)-F(k)]$

and then cancel out terms to get F(n+1)-F(1). Sometimes the desired identity is hard to find, but basically you are searching for it in the recursive form of the sequence, or you can look foor the “conjugates” for the terms you have. The first example is classical. You Certainly know these formulas. $\displaystyle\sum^{n}_{k=1}k=\dfrac{n(n+1)}{2}$ $\displaystyle\sum^{n}_{k=1}k^2=\dfrac{n(n+1)(2n+1)}{6}$ $\displaystyle\sum^{n}_{k=1}k^3=\left[\dfrac{n(n+1)}{2}\right]^2$

What about $\displaystyle\sum^{n}_{k=1}k^4$?

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