Question

by mathematical induction s.t 1square+2sqare+.......+nsquare=n(n+1)(2n+1)/2​

Answer 1

Answers

rishu6845

rishu6845 Genius

Step-by-step explanation:

To prove ---->

--------------

n(n+1)(2n+1)

1²+2²+..................+n²=-----------------------

6

proof---> we prove it by pmi

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first we prove it for n=1

first term=1²=1

now putting n=1 in RHS

1(1+1)(2×1+1)

=---------------------

6

1(2)(2+1)

=--------------------

6

6

=--------=1

6

so given statement is true for n=1

now let given statement is true for n=k

k(k+1)(2k+1)

1²+2²+3²+........+k²=-----------------------

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now we prove it is true for n=k+1

for this adding (k+1)² to both sides

k(k+1)(2k+1)

1²+2²+....+k²+(k+1)²=-------------------- +(k+1)²

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k(2k+1)

=(k+1){--------------------------- +(k+1) }

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k(2k+1) + 6 (k+1)

=(k+1) {_____________________}

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2k²+ k + 6k + 6

= (k+1) (--------------------------------)

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2k²+7k+6

= (k+1) (------------------------)

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2k² +(4+3)k +6

=(k+1) {------------------------------}

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2k²+4k+3k+6

=(k+1) {------------------------------}

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2k(k+2) +3(k+2)

=(k+1) { _________________}

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(k+2) (2k+3)

=(k+1) {________________}

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{(k+1)+1}{(2k+2)+1}

=(k+1) [_____________________]

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(k+1) {(k+1) +1 } {2(k+1)+1}

=--------------------------------------------------

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so given statement is true for n=k+1 if it's true for n=k

so by principle of mathematical induction given statement is true for any natural number

Hope it helps you

Answer 2

$\huge \sf \fcolorbox{red}{pink}{Solution :)}$

Let ,

$\sf P(n) = {(1)}^{2} + {(2)}^{2} + {(3)}^{2} + ..... + {(n)}^{2} = \frac{n(n + 1)(2n + 1)}{6} \: - - - \: eq(i)$

For n = 1 , LHS of equation (i) = 1 and RHS of equation (i) = 6/6 = 1

$\sf \therefore LHS = RHS$

Therefore , P(1) is true

Assume P(K) is true , then

$\sf P(k) = {(1)}^{2} + {(2)}^{2} + {(3)}^{2} + ..... + {(k)}^{2} = \frac{k(k + 1)(2k + 1)}{6}$

For n = K + 1

$\sf P(k) = {(1)}^{2} + {(2)}^{2} + {(3)}^{2} + ..... + {(k)}^{2} + {(k + 1)}^{2} = \frac{(k + 1)(k + 2 )(2k + 3)}{6}$

$\sf \hookrightarrow RHS = \frac{k(k + 1)(2k + 1)}{6} + {(k + 1)}^{2} \\ \\\sf \hookrightarrow RHS = (k + 1) \bigg(\frac{k(2k + 1) + 6(k + 1)}{6} \bigg) \\ \\\sf \hookrightarrow RHS = (k + 1) \bigg(\frac{2 {(k)}^{2} + 7k + 6}{6} \bigg) \\ \\\sf \hookrightarrow RHS = \frac{(k + 1)(k + 2)(2k + 3)}{6} = LHS$

Therefore , P(k+1) is true , hence by PMI , P(n) is true for all natural number

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