By expanding $(r+1)^4 -r^4$ calculate $\sum r^3$
step 1 - expand to get $1 + 4r + 6r^2 + 4r^3 $
step 2 - what should I do next to calculate the sum?
I don't really understand why we have been given $(r+1)^4 -r^4$. Why have we been given this? How does it help in any way?
It helps, via what we would call telescoping. Note the following: $$ (r+1)^4-r^4 = 4r^3 + 6r^2+4r+1 $$ Suppose we are asked to calculate $\sum_{r=1}^n r^3$. We would repeatedly substitute $r = 2,\ldots n$ in the above expression, and sum them up: $$ 2^4 - 1^4 = 4\cdot 1^3 + 6 \cdot 1^2 + 4\cdot r + 1 $$ $$ 3^4 - 2^4 = 4\cdot 2^3 + 6 \cdot 2^2 + 4\cdot 2 + 1 $$ $$ \vdots $$ $$ (n+1)^4 - n^4 = 4\cdot n^3 + 6 \cdot n^2 + 4\cdot n + 1 $$ Add these all up, and see what is happening on the left side: terms are getting cancelled, and you will end up with $(n+1)^4 - 1^4$, and on the right side: $$ 4\sum_{i=1}^n i^3 + 6\sum_{i=1}^n i^2 + 4\sum_{i=1}^n i + n $$ Now, we know that:$$ \sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6} ; \sum_{i=1}^n i = \frac{n(n+1)}{2} $$ Substitute these: $$ (n+1)^4-1 = 4\sum_{i=1}^n i^3 + 6\left(\frac{n(n+1)(2n+1)}{6}\right) + 4\left(\frac{n(n+1)}2\right) + n $$ Simplify: $$ (n+1)^4-1 = 4\sum_{i=1}^n i^3 + n(n+1)(2n+1) + 2n(n+1) + n $$ Expand and collect like terms: $$ 4\sum_{i=1}^n i^3 = n^4+2n^3+n^2 = n^2(n+1)^2 $$ Therefore, you will get: $$ \sum_{i=1}^n i^3 = \left(\frac{n(n+1)}{2}\right)^2 $$ This is a technique, called telescoping, and the above can be extended by induction to larger powers.