Question

Question #81d37

Answer 1

This is a classical case in which you can prove something by induction. The idea is the following: think of a river with very strong current that you must cross. You can't swim, but there are several stones, and you can jump from one to the other. If you find a first (solid!) stone, and you learn how to jump from a stone to the next, then you can cross the river!

In formulas, a proof by induction works as follows:

1. First of all, you must prove that the formula holds for a basic case (which is usually `n=0` or `n=1`)
2. You must prove that, if the formula holds for a generic value `n`, then it also works for `n+1`.

By doing so, you can work -indeed- by induction, and step from a number to its successor, and so cover the entire range of natural numbers: the statement it's true for `n=1`. Then, it's true for `n=2`. Applying the same logic, we can step from `n=2` to `n=3`, and so on.

So, let's begin with the basic case `n=1`. Is it true that

`\sum_{i=1}^1 i^3= {1^2*(1+1)^2}/4`?

Of course it is, since the sum means simply 1^3, which is one, and the fraction on the right side is also 1.

So now, let's assume that the sum of the first `n` cubes is `{n^2*(n+1)^2}/4`, and we must prove that the sum of the first `n+1` cubes is `{(n+1)^2*[(n+1)+1]^2}/4`, which is the previous formula in which I substituted `n\rightarrow n+1`. We have:

`\sum_{i=1}^{n+1} i^3=\sum_{i=1}^{n} i^3 + (n+1)^3`, and we are assuming that the sum up to `n` equals `{n^2*(n+1)^2}/4`. We simply need to add `(n+1)^3` and check the result:

`{n^2*(n+1)^2}/4+(n+1)^3=`
`{n^2 (n^2+2n+1) + 4(n^3+3n^2+3n+1)}/4=`
`{n^4+2n^3+n^2+4n^3+12n^2+12n+4}/4=`
`{n^4+6n^3+13n^2+12n+4}/4`

We want this to be equal to `{(n+1)^2*[(n+1)+1]^2}/4`, which is true if and only if the numerators are equal. Let's expand this last numerator:
`(n+1)^2*[(n+1)+1]^2=`
`(n^2+2n+1)(n+2)^2=`
`(n^2+2n+1)(n^2+4n+4)=`
`n^4+4n^3+4n^2+2n^3+8n^2+8n+n^2+4n+4=`
`n^4+6n^3+13n^2+12n+4`

which is exactly what we wanted.

Answer 2

First of all let's try to understand why it works.

We know that `sum_(i=1)^ni=(n(n+1))/2`,

and this is because we have `n/2` sums of the value `n+1` (that is because the first added to the last, the second added to the second-last, and so on... have all value `n+1`).

Now we can see our sum:

`sum_(i=1)^ni^3=1+8+27+64+125+...+n^3`

Let's proceed step by step:

`S_1=sum_(i=1)^1i^3=1`,

`S_1^'=sum_(i=1)^1i=1`,

so `S_1= (S_1^')^2`.

`S_2=sum_(i=1)^2i^3=1+8=9`,

`S_2^'=sum_(i=1)^2i=1+2=3`,

so `S_2= (S_2^')^2`.

`S_3=sum_(i=1)^3i^3=1+8+27=36`,

`S_3^'=sum_(i=1)^3i=1+2+3=6`,

so `S_3= (S_3^')^2`.

...

`S_n=sum_(i=1)^ni^3=1+8+27+64+125+...+n^3`,

`S_n^'=sum_(i=1)^1i=1+2+3+4+5+...+n=(n(n+1))/2`,

so `S_n` has to be `(S_n^')^2` and so:

`S_n=((n(n+1))/2)^2=(n^2(n+1)^2)/4`.

But this is not a demonstration.

This is only an idea.

We have to demonstrate it with the induction rule, that says:

If a property is true for the first value, we can assume that it is true for the `n`th term and we have to demonstrate that it is true for the `(n+1)`th term.

So:

Is the property true for the first term?

`sum_(i=1)^1i^3=1` and `(1^2(1+1)^2)/4=1`: YES

Let's assume that it is true for `n`:

`sum_(i=1)^ni^3=(n^2(n+1)^2)/4` and this is our hypotesis.

Let's demonstrate that it is true also for `n+1`:

`sum_(i=1)^(n+1)i^3=sum_(i=1)^ni^3+(n+1)^3=`

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

`=((n+1)^2(n^2+4n+4))/4=((n+1)^2(n+2)^2)/4`

that is perfectly the formula that we have to demonstrate in which at the place of `n` there is `n+1`!