1. ## Homework problem

We know that

$\sum_{k=1}^nk=\frac{n(n+1)}{2}$

where $n$ is a natural number. obtain this formula using the identity.

$(k+1)^2-k^2=2k+1$

to obtain the expression

$\sum_{k=1}^n(2k+1)$ then use a similar approach to obtain the formula

$\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{6}$

DO NOT USE INDUCTION

There is the question, I don't even know where to begin :/

2. Originally Posted by jpatrie
We know that

$\sum_{k=1}^nk=\frac{n(n+1)}{2}$

where $n$ is a natural number. obtain this formula using the identity.

$(k+1)^2-k^2=2k+1$

to obtain the expression

$\sum_{k=1}^n(2k+1)$ then use a similar approach to obtain the formula

$\sum_{k=1}^n(2k+1) = \sum_{k=1}^n (k+1)^2 - k^2$

......... $= (2^2-1^2)+(3^2-2^2)+ ... + (n^2-(n-1)^2) + ((n+1)^2-n^2)$

Now all but the $-1^2$ and $(n+1)^2$ terms cancell, so:

$\sum_{k=1}^n(2k+1) =(n+1)^2-1$

RonL

3. Hello, jpatrie!

We know that: . $\sum_{k=1}^nk=\frac{n(n+1)}{2}$, where $n$ is a natural number.

Obtain this formula using the identity: . $(k+1)^2-k^2\:=\:2k+1$

We have: . $(k+1)^2 - k^2 \:=\:2k+1$

Let $k = 1,2,3,\hdots, n$ and "stack" the equations.

. . $\begin{array}{cccc} k=1\!: & 2^2 - 1^2 &=& 2(1) + 1 \\ k=2\!: & 3^2 - 2^2 &=& 2(2) + 1 \\ k=3\!: & 4^2 - 3^2 &=& 2(3) + 1 \\ \vdots & \vdots & & \vdots \\ k=n\!: & (n+1)^2 - n^2 &=& 2n + 1 \end{array}$

Now add the equations ... note that most of the left side cancels out.

. . $(n+1)^2 - 1 \;=\;2\underbrace{(1 + 2 + 3 + \hdots + n)}_{\text{This is }\sum k} + \underbrace{(1 + 1 + 1 + \hdots + 1)}_{n\text{ terms}}$

So we have: . $n^2 +2n \;=\;2\sum k + n \quad\Rightarrow\quad 2\sum k \;=\;n^2 + n$

Therefore: . $\boxed{\sum k \;=\;\frac{n(n+1)}{2}}$

Then use a similar approach to obtain the formula: . $\sum_{k=1}^nk^2\;=\;\frac{n(n+1)(2n+1)}{6}$

We have: . $(k+1)^3 - k^3 \;=\;3k^2 + 3k + 1$

Let $k = 1,2,3,\cdots, n$

. . $\begin{array}{ccccc}k=1\!: & 2^3 - 1^3 &=& 3(1^2) + 3(1) + 1 \\
k=2\!: & 3^3 - 2^3 &=& 3(2^2) + 3(2) + 1 \\
k=3\!: & 4^3 - 3^3 &=& 3(3^2) + 3(3) + 1 \\
\vdots & \vdots & & \vdots \\
k=n\!: & (n+1)^3 - n^3 &=& 3(n^2) + 3(n) + 1 \end{array}$

. . $(n+1)^3 - 1 \;=\;3\underbrace{\left[1^2 + 2^2 + 3^2 +\hdots+n^2\right]}_{\text{This is }\sum k^2} \:+\: 3\underbrace{\left[1 + 2 + 3 +\hdots+n\right]}_{\text{This is }\sum k} +$ . $\underbrace{[1 + 1 + 1 + \hdots + 1]}_{n\text{ terms}}$

So we have: . $n^3 + 3n^2 + 3n \;=\;3\sum k^2 + 3\cdot\frac{n(n+1)}{2} + n$

. . which simplifies to: . $3\sum k^2 \;=\;\frac{2n^3 + 3n^2 + n}{2} \;=\;\frac{n(n+1)(2n+1)}{2}$

Therefore: . $\boxed{\sum k^2 \;=\;\frac{n(n+1)(2n+1)}{6}}$

4. Thanks a lot guys, but can anyone prove it using "recursion", unless those are the methods of recursion.