Question

How do you evaluate the sum represented by ∑ with n^3 ? Any examples

Answer 1

See example below.

Explanation:

Example

`sum_(n=1)^5 n^3 = 1^3+2^3+3^3+4^3+5^3`

`= 1+8+27+64+125 = 225`

Answer 2

Amazingly, you can also say that `sum_{n=1}^{k}n^3=1^3+2^3+3^3+4^3+cdots+k^3=k^4/4+k^3/2+k^2/4=((k^2+k)^2)/4` for any integer `k geq 1`.

Explanation:

Deriving this equation in the first place is tricky. But proving it is not too hard (when you are told what the answer is) if you know how to use the principle of mathematical induction .

The base case is the `k=1` case, which is easy: `sum_{n=1}^{1}n^3=1^3=1` while `1^4/4+1^3/2+1^2/4=1/4+1/2+1/4=1`.

Now assume that `sum_{n=1}^{k}n^3=k^4/4+k^3/2+k^2/4` for some fixed integer `k geq 1`. The principle of mathematical induction says that if you can now prove this last equation is true when `k` is replaced by `k+1`, the equation will be true for all integers `k geq 1`.

In other words, to finish the proof, we must use our assumption above to prove that `sum_{n=1}^{k+1}n^3=(k+1)^4/4+(k+1)^3/2+(k+1)^2/4`.

This can be done by doing the following algebraic steps (our assumption above is used in the 2nd equality):

`sum_{n=1}^{k+1}n^3=(k+1)^3+sum_{n=1}^{k}n^3`

`=(k+1)^3+k^4/4+k^3/2+k^2/4`

`=k^3+3k^2+3k+1+k^4/4+k^3/2+k^2/4`

`=1/4 k^4+3/2 k^3+13/4 k^2+3k+1`,

while

`(k+1)^4/4+(k+1)^3/2+(k+1)^2/4`

`=1/4 (k^4+4k^3+6k^2+4k+1)+1/2 (k^3+3k^2+3k+1)+1/4 (k^2+2k+1)`

`=1/4 k^4+k^3+3/2 k^2+k+1/4+1/2 k^3+3/2 k^2+3/2 k+1/2+1/4 k^2+1/2 k+1/4`

`=1/4 k^4+3/2 k^3+13/4 k^2+3k+1`.

Therefore, by the principle of mathematical induction, `sum_{n=1}^{k}n^3=k^4/4+k^3/2+k^2/4=((k^2+k)^2)/4` for any integer `k geq 1`.

Answer 3

Derive the formula for `sum_(k=1)^n k^3` by evaluating differences of sequences.

Explanation:

One way of deriving the formula that Bill gives is as follows:

The first few sums are:

`sum_(k=1)^1 k^3 = 1`

`sum_(k=1)^2 k^3 = 1+8 = 9`

`sum_(k=1)^3 k^3 = 9+27 = 36`

`sum_(k=1)^4 k^3=36+64=100`

`sum_(k=1)^5 k^3=100+125=225`

`sum_(k=1)^6 k^3=225+216=441`

`sum_(k=1)^7 k^3=441+343=784`

I know I don't need the last two, since the formula for the sum will have highest order term proportional to `n^4`, but they help you see the way the following becomes a constant sequence...

Write these values as a sequence, then form the sequence of differences and repeat to get:

`color(blue)(1), 9, 36, 100, 225, 441, 784`

`color(blue)(8), 27, 64, 125, 216, 343`

`color(blue)(19), 37, 61, 91, 127`

`color(blue)(18), 24, 30, 36`

`color(blue)(6), 6, 6`

Having reached a constant sequence, read off the first term of each sequence to give coefficients:

`sum_(k=1)^n k^3 = color(blue)(1)/(0!) + color(blue)(8)/(1!)(n-1) + color(blue)(19)/(2!)(n-1)(n-2) + color(blue)(18)/(3!)(n-1)(n-2)(n-3) + color(blue)(6)/(4!)(n-1)(n-2)(n-3)(n-4)`

`=1+8n-8+19/2n^2-57/2n+19+3n^3-18n^2+33n-18+1/4n^4-5/2n^3+35/4n^2-25/2n+6`

`=n^2/4+n^3/2+n^4/4`