Easy Math question : Sum of squares

Question

How to sum $2^2 + 4^2 + 6^2 + \dots + (2n)^2$ upto n terms. Also what if we have to sum $1^2 + 3 ^2 + \dots + (2n+1)^2$ up to n terms.

I am new to this topic so please answer in a simple manner

Answer 1

We have the formula $$1^2+2^2+3^2+4^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}$$

Multiplying both sides by $4$:

$$2^2+4^2+\cdots+(2n)^2=\frac{2n(n+1)(2n+1)}{3}$$

This gives your first sum.

For the second, substitute $n=2n_1$ in the first equation,

$$1^2+2^2+\cdots+(2n)^2=\frac{2n(2n+1)(4n+1)}{6}$$

Subtracting the answer to the first part, from this new equation gives

$$1^2+3^2+\cdots+(2n-1)^2=\frac{2n(2n+1)(4n+1)}{6}-\frac{2n(n+1)(2n+1)}{3}$$

Simply add $(2n+1)^2$ to both sides.

Answer 2

Hint: You might know, $1^2+2^2+3^2+\ldots+n^2=\dfrac{n(n+1)(2n+1)}{6}$

$2^2+4^2+6^2+\ldots+(2n)^2=4(1^2+2^2+\ldots+n^2)$

Can you extend this to answer other part of your question?

Answer 3

$$2^2+4^2+6^2+\cdots +(2n)^2$$

$$=2^2\cdot1+2^2\cdot2^2+2^2\cdot 3^2+\cdots+2^2\cdot n^2$$

$$=2^2\cdot(1^2+2^2+\cdots+n^2)$$

Can you conclude now?

Once you Know what $2^2+4^2+6^2+\cdots +(2n)^2$ is..

It would not take much time to see what $1^2+3^2+\cdots+(2n-1)^2$ is as :

$$[1^2+3^2+\cdots+(2n-1)^2]+[2^2+4^2+6^2+\cdots +(2n)^2]=1^2+2^2+\cdots+(2n)^2$$