Solveeit Logo

Question

Question: If the sum of first \(n\) natural number is one-fifth of the sum of their squares, then \(n\) is, ...

If the sum of first nn natural number is one-fifth of the sum of their squares, then nn is,
(a) 55
(b) 66
(c) 77
(d) 88

Explanation

Solution

Hint: Use the formulas of sequences and series to find the sum of nn natural numbers and their squares. Substitute these formulas in the equation which can be obtained by reading the question and then find the value of nn.

Complete step-by-step answer:

Before proceeding with the question, we must know all the formulas of the sequences and series which will be required to solve this question.
We have a formula from which, the sum of first rr natural numbers (denoted by r\sum{r}) is given by,
r=r(r+1)2....................(1)\sum{r=\dfrac{r\left( r+1 \right)}{2}}....................\left( 1 \right)
Also, we have a formula from which, the sum of the squares first rr natural number (denoted by r2\sum{{{r}^{2}}}) is given by,
r2=r(r+1)(2r+1)6................(2)\sum{{{r}^{2}}=\dfrac{r\left( r+1 \right)\left( 2r+1 \right)}{6}................\left( 2 \right)}
In the question, it is given that the sum of the first nn natural number is one-fifth of the sum of their squares.
\Rightarrow sum of first nn natural numbers == 15×\dfrac{1}{5}\times sum of squares of first nn natural numbers
\Rightarrow n=15n2.................(3)\sum{n}=\dfrac{1}{5}\sum{{{n}^{2}}.................\left( 3 \right)}
Substituting r=nr=n in formula (1)\left( 1 \right), the sum of first nn natural numbers is equal to,
n=n(n+1)2....................(4)\sum{n=\dfrac{n\left( n+1 \right)}{2}}....................\left( 4 \right)
Substituting r=nr=n in formula (2)\left( 2 \right), the sum of the squares of first nn natural numbers is equal to,
n2=n(n+1)(2n+1)6................(5)\sum{{{n}^{2}}=\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}................\left( 5 \right)}
Substituting n\sum{n} from equation (4)\left( 4 \right) and n2\sum{{{n}^{2}}} from equation (5)\left( 5 \right) in equation (3)\left( 3 \right), we get,
n(n+1)2=15(n(n+1)(2n+1)6)\dfrac{n\left( n+1 \right)}{2}=\dfrac{1}{5}\left( \dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6} \right)
Cancelling n(n+1)n\left( n+1 \right) on both the sides of the above equation, we get,
12=(2n+1)30 15=2n+1 2n=14 n=7 \begin{aligned} & \dfrac{1}{2}=\dfrac{\left( 2n+1 \right)}{30} \\\ & \Rightarrow 15=2n+1 \\\ & \Rightarrow 2n=14 \\\ & \Rightarrow n=7 \\\ \end{aligned}
Hence, the answer is option (c).

Note: There is a possibility that one may make a mistake while applying the formula for sum of the first nn natural number. It is a very common mistake that one uses the formula as n=n(n1)2\sum{n=\dfrac{n\left( n-1 \right)}{2}} instead of the formula n=n(n+1)2\sum{n=\dfrac{n\left( n+1 \right)}{2}}. This may lead us to an incorrect answer.