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Sunny_sXe [5.5K]
2 years ago
8

How can i prove this property to be true for all values of n, using mathematical induction.

Mathematics
1 answer:
chubhunter [2.5K]2 years ago
8 0

Proof -

So, in the first part we'll verify by taking n = 1.

\implies \: 1  =  {1}^{2}  =  \frac{1(1 + 1)(2 + 1)}{6}

\implies{ \frac{1(2)(3)}{6} }

\implies{ 1}

Therefore, it is true for the first part.

In the second part we will assume that,

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

and we will prove that,

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

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

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

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

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

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

{1}^{2} +  {2}^{2}  +  {3}^{2}  + ..... +  {k}^{2}  + (k + 1)^{2} = \frac{(k+1)(2k^2+7k+6) }{6}

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

<u>Henceforth, by </u><u>using </u><u>the </u><u>principle </u><u>of </u><u> mathematical induction 1²+2² +3²+....+n² = n(n+1)(2n+1)/ 6 for all positive integers n</u>.

_______________________________

<em>Please scroll left - right to view the full solution.</em>

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