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 n natural number is one-fifth of the sum of their squares, then n is,
(a) 5
(b) 6
(c) 7
(d) 8
Solution
Hint: Use the formulas of sequences and series to find the sum of n 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 n.
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 r natural numbers (denoted by ∑r) is given by,
∑r=2r(r+1)....................(1)
Also, we have a formula from which, the sum of the squares first r natural number (denoted by ∑r2) is given by,
∑r2=6r(r+1)(2r+1)................(2)
In the question, it is given that the sum of the first n natural number is one-fifth of the sum of their squares.
⇒ sum of first n natural numbers = 51× sum of squares of first n natural numbers
⇒ ∑n=51∑n2.................(3)
Substituting r=n in formula (1), the sum of first n natural numbers is equal to,
∑n=2n(n+1)....................(4)
Substituting r=n in formula (2), the sum of the squares of first n natural numbers is equal to,
∑n2=6n(n+1)(2n+1)................(5)
Substituting ∑n from equation (4) and ∑n2 from equation (5) in equation (3), we get,
2n(n+1)=51(6n(n+1)(2n+1))
Cancelling n(n+1) on both the sides of the above equation, we get,
21=30(2n+1)⇒15=2n+1⇒2n=14⇒n=7
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 n natural number. It is a very common mistake that one uses the formula as ∑n=2n(n−1) instead of the formula ∑n=2n(n+1). This may lead us to an incorrect answer.