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

I want to see the solution

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Take me to next question

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is B
Take me to next question

For any integer $n \geq 1$ , let $p_{n}$ be the statement that

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

$B a s e c a s e - --------- -$ : The statement $P_{1}$ says that

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

This is true.

$I n d u c t i v e s t e p - -------------- -$ : Fix $k \geq 1$ , and suppose that $P_{k}$ holds, that is,

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Therefore $P_{k + 1}$ holds

Thus, by the principle of mathematical induction, for all $n \geq 1$ , $P_{n}$ holds.

Suggest Corrections

Similar questions

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

If r is a fixed positive integer, prove by induction that (r+1)(r+2)(r+3)....(r+n) is divisible by n!

Prove using mathematical induction that for all $n \geq 1$

1+4+7+..+(3n-2)= $\frac{n (3 n - 1)}{2}$

Q. Prove the following by using the principle of mathematical induction for all

n \in N : 1^{2} + 3^{2} + 5^{2} + . . . . . . . + ({2 n - 1)}^{2} = \frac{n (2 n - 1) (2 n + 1)}{3}

Prove the following by using the principle of mathematical induction for all n ∈ N:

Join BYJU'S Learning Program

Verify that for all n≥1 , the sum of the squares of the first 2n positive integers is given by the formula 12+22+32+.....(2n)2=n(2n+1)(4n+1)3

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