For all n - 1- prove that 12 - 22 - 32 - 42 - - - n2 - nn - 12n - 16

Let P( n ) be the given statement. P( n ) = 1^2 + 2^2 + 3^2 + #160; ⋯ #160; + n^2 = n( n + 1)( 2n + 1)6 By principle of mathematical induction, le ...

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Standard XII

Mathematics

Proof by mathematical induction

For all n ≥...

Question

For all $n \geq 1,$ prove that
$1^{2} + 2^{2} + 3^{2} + 4^{2} + . . . . + n^{2} = \frac{n (n + 1) (2 n + 1)}{6}$

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Solution

Let $P (n)$ be the given statement.
$P (n) = 1^{2} + 2^{2} + 3^{2} + \dots + n^{2} = \frac{n (n + 1) (2 n + 1)}{6}$
By principle of mathematical induction, let the statement is true for $n = 1$ .
$P (1) = 1^{2} = \frac{1 (1 + 1) (2 (1) + 1)}{6}$
$1 = 1$
So, the statement is true for $n = 1$ .
Let the statement is true for $n = k$ , then,
$P (k) = 1^{2} + 2^{2} + 3^{2} + \dots + k^{2} = \frac{k (k + 1) (2 k + 1)}{6}$ (1)
For $n = k + 1$ ,
$1^{2} + 2^{2} + 3^{2} + \dots + {(k + 1)}^{2} = \frac{(k + 1) (k + 2) (2 k + 3)}{6}$ (2)
The left hand side of the above equation can be written as,
$1^{2} + 2^{2} + 3^{2} + \dots + k^{2} + {(k + 1)}^{2}$
From equation (1),
$1^{2} + 2^{2} + 3^{2} + \dots + k^{2} + {(k + 1)}^{2} = \frac{k (k + 1) (2 k + 1)}{6} + {(k + 1)}^{2}$
$= \frac{k (k + 1) (2 k + 1) + 6 {(k + 1)}^{2}}{6}$
$= \frac{(k + 1) (k (2 k + 1) + 6 (k + 1))}{6}$
$= \frac{(k + 1) (2 k^{2} + 7 k + 6)}{6}$
$= \frac{(k + 1) (k + 2) (2 k + 3)}{6}$
This is equal to the left hand side of the equation (2).
Therefore, by the principle of mathematical induction, the given statement is true for every real number.

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Similar questions

Q. For all

n \geq 1

, prove that

1^{2} + 2^{2} + 3^{2} + 4^{2} + \dots + n^{2} = \frac{n (n + 1) (2 n + 1)}{6}

Q. Prove by Mathematical Induction, for all

n \in N

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

Q. Prove by method of induction, for all

n \in

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

Q. Prove by mathematical induction,

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

Q. Show that

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

for all

n \in N

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For all n≥1, prove that 12+22+32+42+....+n2=n(n+1)(2n+1)6

For all $n \geq 1,$ prove that
$1^{2} + 2^{2} + 3^{2} + 4^{2} + . . . . + n^{2} = \frac{n (n + 1) (2 n + 1)}{6}$