Question

this one divided by 6

Answer 1

Hope you like my process
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$= > p(n) = {1}^{2} + {2}^{2} + {3}^{2} + .... + {n}^{2} = \frac{n(n + 1)(2n + 1)}{6} \\ \\ = > p(1) = {1}^{2} = \frac{1(1 + 1)(2 + 1)}{6} = \frac{2 \times 3}{6} = 1 \\ \\ \bf \: \: so \: \: p(1) \: \: is \: \: true \: \\ \\ = > let \: \: p(k) \: \: be \: true \\ \\ = > \bf \: p(k) = {1}^{2} + {2}^{2} + {3}^{2} + ... + {k}^{2} = \frac{k(k + 1)(2k + 1)}{6} \\ \\ = > { \bf \: p(k + 1)} = {1}^{2} + {2}^{2} + {3}^{2} + ... + {k}^{2} + {(k +1 )}^{2} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{k(k + 1)(2k + 1)}{6} + {(k + 1)}^{2} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = { \bf(k + 1)}{( \frac{k(2k + 1)}{6} + (k + 1)) } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = { \bf(k + 1)}( \frac{2 {k}^{2} + k + 6(k + 1)}{6} ) \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = { \bf(k + 1)}( \frac{2 {k}^{2} + k + 6k + 6 }{6} ) \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{(k + 1)}{6} (2 {k}^{2} + 4k + 3k + 6) \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{(k + 1)}{6} (2k(k + 2) + 3(k + 2)) \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{(k + 1)(k + 2)(2k + 3)}{6} = \bf{ \underline{ \blue{ \frac{(k + 1)((k + 1) + 1)(2(k + 1) + 1)}{6} }}}$

Thus P(k) and P(k+1) is true .

Hence P(n) is true for all natural numbers n€N
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Hope this is your required answer

Proud to help you

Answer 2

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