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Question: It is given that the sum of n terms \(\sum {n = 55} \) then, what is the value of \(\sum {{n^2}} \) ...

It is given that the sum of n terms ∑n=55\sum {n = 55} then, what is the value of ∑n2\sum {{n^2}} ?

A. 385 B. 506 C. 1115 D. 3025  A.{\text{ }}385 \\\ B.{\text{ }}506 \\\ C.{\text{ }}1115 \\\ D.{\text{ }}3025 \\\
Explanation

Solution

Hint: Since the sum of first n natural number is given by n(n+1)2\dfrac{{n(n + 1)}}{2} , with the help of this calculate the value of n and then put it in the formula ∑n2=n(n+1)(2n+1)6\sum {{n^2} = \dfrac{{n(n + 1)(2n + 1)}}{6}} to calculate the sum of square of n terms.

Given that:
∑n=55\sum {n = 55} …………………. (1)
We know that sum of first n natural number
=n(n+1)2= \dfrac{{n(n + 1)}}{2} ……………………. (2)
Now, equating equation 1 with 2 to get the value of n
⇒n(n+1)2=55 ⇒n2+n=110 ⇒n2+n−110=0  \Rightarrow \dfrac{{n(n + 1)}}{2} = 55 \\\ \Rightarrow {n^2} + n = 110 \\\ \Rightarrow {n^2} + n - 110 = 0 \\\
Solving the quadratic equation, we get
⇒n2−10n+11n−110=0 ⇒(n+11)(n−10)=0 ⇒n=10  \Rightarrow {n^2} - 10n + 11n - 110 = 0 \\\ \Rightarrow (n + 11)(n - 10) = 0 \\\ \Rightarrow n = 10 \\\
Neglecting the negative value of n because n is a natural number.

Using the formula to calculate sum of n square terms
∑n2=n(n+1)(2n+1)6\sum {{n^2} = \dfrac{{n(n + 1)(2n + 1)}}{6}}
Putting the value of n in this formula, we get
=10(10+1)(2×10+1)6 =10×11×216 =385  = \dfrac{{10(10 + 1)(2 \times 10 + 1)}}{6} \\\ = \dfrac{{10 \times 11 \times 21}}{6} \\\ = 385 \\\
Hence, the sum of squares of n terms is 385.

Option A is the correct option.

Note: To solve these types of series problems, remember the formula of sum of n natural numbers, sum of square of n natural numbers and sum of square of cube of n natural numbers. After this see the conditions given in the question and then see number of unknown variables is equal to number of equations, then start solving for unknown variables.