For what value of n will the sum of natural numbers -1-5 the sum of squares of the natural numbers-

Sum of n natural number is n(n+1)/2 And square of n natural number is n(n+1)(2n+1)/6 So by condition, we have n(n+1)/2=1/5[n(n+1)(2n+1)/6] Or 15 = 2n+1 n=7 ...

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Standard IX

Mathematics

Method of Elimination

For what valu...

Question

For what value of n will the sum of natural numbers = $\frac{1}{5}$ the sum of squares of the natural numbers.

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Solution

Sum of n natural number is $\frac{n (n + 1)}{2}$
And square of n natural number is $\frac{n (n + 1) (2 n + 1)}{6}$
So by condition, we have
$\frac{n (n + 1)}{2} = \frac{1}{5} [\frac{n (n + 1) (2 n + 1)}{6}]$
Or 15 = 2n+1
n=7

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For what value of n will the sum of natural numbers = 15 the sum of squares of the natural numbers.

Sum of n natural number is n(n+1)2 And square of n natural number is n(n+1)(2n+1)6 So by condition, we have n(n+1)2=15[n(n+1)(2n+1)6] Or 15 = 2n+1 n=7

For what value of n will the sum of natural numbers = $\frac{1}{5}$ the sum of squares of the natural numbers.

Sum of n natural number is $\frac{n (n + 1)}{2}$
And square of n natural number is $\frac{n (n + 1) (2 n + 1)}{6}$
So by condition, we have
$\frac{n (n + 1)}{2} = \frac{1}{5} [\frac{n (n + 1) (2 n + 1)}{6}]$
Or 15 = 2n+1
n=7