Sum it up!

Algebra Level 2

$\sum_{k=1}^5 (k^2 + 2)$ .

The answer is 65.

1 solution

Prasun Biswas
Apr 6, 2014

We use here the formula $1^2+2^2+3^2+.....+n^2=\frac{1}{6}n(n+1)(2n+1)$ and the property of summation which suggests that $\displaystyle \sum_{i=a}^n(x+y)=\sum_{i=a}^n(x)+\sum_{i=a}^n(y)$

$\displaystyle \sum_{k=1}^5(k^2+2)$

$=\displaystyle \sum_{k=1}^5(2)+ \sum_{k=1}^5(k^2)$

$=(2+2+2+2+2)+(1^2+2^2+3^2+4^2+5^2)$

$=10+(\frac{1}{6}\times 5\times (5+1)\times ((2\times 5) + 1))$

$=10+(\frac{1}{6}\times 5\times 6\times (10+1))$

$=10+(\frac{1}{6}\times 5\times 6\times 11)$

$= 10+(5\times 11)=10+55=\boxed{65}$

This method will allow you to solve this problem even when the limits of the summation are from $k=1$ to any large value of $k$ as simple summing up takes up a long amount of time when limits of sum are large... :)

Prasun Biswas - 7 years, 2 months ago

Just asking, where did you get this formula?

Shabarish Ch - 7 years, 2 months ago

And if there are more such formulas, where can I know about them?

Shabarish Ch - 7 years, 2 months ago

You can google it but you can also try to derive them by yourselves. You will learn these series sum formulas in class 11 in the chapter "Sequence and Series" of the NCERT Maths book of class 11. Here are some formulas that I have in mind right now ---->

$1+2+3+.....+n=\frac{1}{2}n(n+1)$

$1^2+2^2+3^2+......+n^2=\frac{1}{6}n(n+1)(2n+1)$

$1^3+2^3+3^3+.....+n^3=(\frac{1}{2}n(n+1))^2$

$1+3+5+......+(2n-1)=n^2$

There are many other such formulas that can be derived from basic math properties but I am too lazy to write the derivations here.... :D :P

Prasun Biswas - 7 years, 2 months ago

@Prasun Biswas – Thank you very much for the info. I will surely look it up...

Shabarish Ch - 7 years, 2 months ago

$\displaystyle \sum^{a}_{b} c$ , so sum function(or sigma function) is too hard for me.

. . - 3 months, 3 weeks ago

Sum it up!

The answer is 65.

1 solution

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