Question

Sigma n =210 then find sigma n^2

Answer 1

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Answer 2

answer is 2870

it is given that, $\Sigma{n}=210$

and we have to find $\Sigma{n^2}$

actually $\Sigma{n}$ is general form of sum of n natural numbers.

i.e., $\Sigma{n}$ = 1 + 2 + 3 + 4 + 5 + .... + n

⇒210 = n(n + 1)/2 [ sum of n natural number is n(n+1)/2 ]

⇒ 420 = n(n + 1)

⇒420 = n² + n

⇒n² + 21n - 20n - 420 = 0

⇒n(n + 21) - 20(n + 21) = 0

⇒(n - 20)(n + 21) = 0

⇒n = 20, -21

hence, there are 20 natural numbers.

so now $\Sigma{n^2}$ = 1² + 2² + 3² + 4² + .... + 20²

= 20(20 + 1)(2 × 20 + 1)/6 [ sum of square of n natural numbers is n(n + 1)(2n + 1)/6 ]

= 20 × 21 × 41/6

= 420 × 41/6

= 70 × 41

= 2870

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