Question

Verify that for all $n\ge 1$ , the sum of the squares of the first 2n positive integers is given by the formula $1^2+2^2+3^2+.....(2n{)}^2=\frac{n(2n+1)(4n+1)}{3}$

• A. I want to see the solution
• B. Take me to next question

Answer 1

The correct option is B

Take me to next question

For any integer $n\ge 1$ , let $p_n$ be the statement that

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

$\underset{–}{Basecase}$ : The statement $P_1$ says that

$1^2+2^2=\frac{\left(1\right)\left(2\right(1)+1)\left(4\right(1)+1)}{3}=\frac{3\left(5\right)}{3}=5$,

This is true.

$\underset{–}{Inductivestep}$: Fix $k\ge 1$ , and suppose that $P_k$ holds, that is,

$1^2+2^2+3^2+.....+(2k{)}^2=\frac{k(2k+1)(4k+1)}{3}$.

It remains to show that $P_{k+1}$ holds, that is,

$1^2+2^2+3^2+.....+\left(2\right(k+1){)}^2=\frac{(k+1)\left(2\right(k+1)+1)\left(4\right(k+1)+1)}{3}$.

$1^2+2^2+3^2+.....+\left(2\right(k+1){)}^2=1^2+2^2+3^2+.....(2k+2{)}^2$

=$1^2+2^2+3^2+.....+(2k{)}^2+(2k+1{)}^2+(2k+2{)}^2$

=$\frac{k(2k+1)(4k+1)}{3}+(2k+1{)}^2+(2k+2{)}^2$ $\left(by{P}_k\right)$

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

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

$=\frac{k(8k^2+6k+1)+3(4k^2+4k+1)+3(4k^2+8k+4)}{3}$

$=\frac{(8k^3+6k^2+k)+(12k^2+12k+3)+(12k^2+24k+12)}{3}$

$=\frac{8k^3+30k^2+37k+15}{3}$

on the otherside of $P_{k+1}$,

$=\frac{(k+1)\left(2\right(k+1)+1)\left(4\right(k+1)+1)}{3}=\frac{(k+1)(2k+2+1)(4k+4+1)}{3}$

= $\frac{(k+1)(2k+3)(4k+5)}{3}$

= $\frac{(2k^2+5k+3)(4k+5)}{3}$

$=\frac{8k^3+30k^2+37k+15}{3}$.

Therefore $P_{k+1}$ holds

Thus, by the principle of mathematical induction, for all $n\ge 1$ ,$P_n$ holds.