1. ## summation of progressions

I have bunch of these kind of tasks at school but i dont know where to start. If someone here could show me how to complete tasks like that so i could do other tasks too.

Prove that, 1x2+2x3+3x4+...+n(n+1)=n(n+1)(n+2)
------------
3

2. ## Re: Some stuff

Originally Posted by Trolo
I have bunch of these kind of tasks at school but i dont know where to start. If someone here could show me how to complete tasks like that so i could do other tasks too.

Prove that, 1x2+2x3+3x4+...+n(n+1)=N(N+1)(N+2)
------------
3
Am I correct in assuming that n and N are the same number? If so, use the same symbol please.

3. ## Re: Some stuff

Do you know the following formulas?
1)$\displaystyle \sum_{i=1}^{n}i=\frac{n(n+1)}{2}$
2)$\displaystyle \sum_{i=1}^{n}i^2=\frac{n(n+1)(2n+1)}{6}$

Note That: $\displaystyle 1 \cdot 2 +2 \cdot 3+3 \cdot 4+...+n(n+1)=\sum_{i=1}^{n}i(i+1)$

$\displaystyle =\sum_{i=1}^{n}(i^2+i)=\sum_{i=1}^{n}i^2+\sum_{i=1 }^{n}i$

4. ## Re: Some stuff

No, ill search for it. Thanks

5. ## Re: Some stuff

in case, you don't find it....

$\displaystyle \text{Prove}$ $\displaystyle \sum_{i=1}^{n}i=\frac{n(n+1)}{2}$
$\displaystyle \sum_{i=1}^{n}i=\underbrace{1+2+3+4+5+6+7+8+...+n} _{\text{This is an arithmetic progression with d=1 and a=1.}}$

Therefore:$\displaystyle \sum_{i=1}^{n}i=\frac{n}{2}{\left[ 2+(n-1)\right]}=\frac{n(n+1)}{2}$

$\displaystyle \text{Prove}$$\displaystyle \sum_{i=1}^{n}i^2=\frac{n(n+1)(2n+1)}{6}$
Consider the identity: $\displaystyle (x+1)^3-x^3=3x^2+3x+1$

Putting $\displaystyle x=1 &,2 &,3 &, ......(n-1) &, n$ we get:

$\displaystyle \boxed{\begin{matrix} 2^3 &-1^3&=&3\cdot1^2 & +3\cdot 1 &+1 \\ 3^3 &-2^3&=&3\cdot2^2 & +3\cdot 2 &+1 \\4^3 &-3^3&=&3\cdot3^2 & +3\cdot 3 &+1 \\ ..&..&&..&..&.. \\ ..&..&&..&..&.. \\ n^3&-(n-1)^3&=&3\cdot (n-1)^2&+3\cdot (n-1)&+1 \\ (n+1)^3&-(n)^3&=&3\cdot (n)^2&+3\cdot (n)&+1 \end{matrix}}$

$\displaystyle (n+1)^3-1^3=3{\left(\sum_{i=1}^{n}i^2 \right)}+3{\left(\sum_{i=1}^{n}i \right)}+n$
$\displaystyle \implies (n+1)^3-1^3=3{\left(\sum_{i=1}^{n}i^2 \right)}+3{\left \{\frac{n(n+1)}{2} \right \}}+n$
$\displaystyle \implies \sum_{i=1}^{n}i^2=\frac{n(n+1)(2n+1)}{6}$