Answer to Question #122823 in Combinatorics | Number Theory for Ojugbele Daniel

Question

Find the sum of the series 1^2 + 2^2 + 3^2 + 4^2 +.... Up to 2n terms.

Answer 1

We will use the Principle of Mathematical Induction to show that $1^2+2^2+3^2+4^2+...+(2n)^2=\frac{n(2n+1)(4n+1)}{3}$

1) Show that it is true for $n=1:$ $1^2+2^2=5=\frac{1\times 3 \times 5}{3}$

2) Assume it is true for $n=k:$ $1^2+2^2+...+(2k)^2=\frac{k(2k+1)(4k+1)}{3}$

Now, prove it is true for $k+1:$

$1^2+2^2+...+(2k+2)^2=(1^2+...+(2k)^2)+(2k+1)^2+(2k+2)^2= \frac{k(2k+1)(4k+1)}{3}+(2k+1)^2+(2k+2)^2= \frac{k(2k+1)(4k+1)+3(2k+1)^2+12(k+1)^2}{3}=\frac{8k^3+30k^2+37k+15}{3}=\frac{(k+1)(2(k+1)+1)(4(k+1)+1)}{3}$

It is true.

So, $1^2+2^2+3^2+4^2+...+(2n)^2=\frac{n(2n+1)(4n+1)}{3}.$