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)
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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)
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3
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: $\displaystyle 1 \cdot 2 +2 \cdot 3+3 \cdot 4+...+n(n+1)=\sum_{i=1}^{n}i(i+1)$
$\displaystyle =\sum_{i=1}^{n}(i^2+i)=\sum_{i=1}^{n}i^2+\sum_{i=1 }^{n}i$
in case, you don't find it....
$\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.}}$$\displaystyle \text{Prove}$ $\displaystyle \sum_{i=1}^{n}i=\frac{n(n+1)}{2}$
Therefore:$\displaystyle \sum_{i=1}^{n}i=\frac{n}{2}{\left[ 2+(n-1)\right]}=\frac{n(n+1)}{2}$
Consider the identity: $\displaystyle (x+1)^3-x^3=3x^2+3x+1$$\displaystyle \text{Prove}$$\displaystyle \sum_{i=1}^{n}i^2=\frac{n(n+1)(2n+1)}{6}$
Putting $\displaystyle x=1 &,2 &,3 &, ......(n-1) &, n $ we get:
$\displaystyle \boxed{\begin{matrix} 2^3 &-1^3&=&3\cdot1^2 & +3\cdot 1 &+1 \\ 3^3 &-2^3&=&3\cdot2^2 & +3\cdot 2 &+1 \\4^3 &-3^3&=&3\cdot3^2 & +3\cdot 3 &+1 \\ ..&..&&..&..&.. \\ ..&..&&..&..&.. \\ n^3&-(n-1)^3&=&3\cdot (n-1)^2&+3\cdot (n-1)&+1 \\ (n+1)^3&-(n)^3&=&3\cdot (n)^2&+3\cdot (n)&+1 \end{matrix}}$
Adding column wise, we obtain:
$\displaystyle (n+1)^3-1^3=3{\left(\sum_{i=1}^{n}i^2 \right)}+3{\left(\sum_{i=1}^{n}i \right)}+n$
$\displaystyle \implies (n+1)^3-1^3=3{\left(\sum_{i=1}^{n}i^2 \right)}+3{\left \{\frac{n(n+1)}{2} \right \}}+n$
Arranging things a little bit...
$\displaystyle \implies \sum_{i=1}^{n}i^2=\frac{n(n+1)(2n+1)}{6}$