Question

It is given that the sum of n terms $\sum {n = 55} $ then, what is the value of $\sum {{n^2}} $ ?
\[
  A.{\text{ }}385 \\
  B.{\text{ }}506 \\
  C.{\text{ }}1115 \\
  D.{\text{ }}3025 \\
 \]

Answer 1

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

Given that:
$\sum {n = 55} $ …………………. (1)
We know that sum of first n natural number
$ = \dfrac{{n(n + 1)}}{2}$ ……………………. (2)
Now, equating equation 1 with 2 to get the value of n
$
   \Rightarrow \dfrac{{n(n + 1)}}{2} = 55 \\
   \Rightarrow {n^2} + n = 110 \\
   \Rightarrow {n^2} + n - 110 = 0 \\
 $
Solving the quadratic equation, we get
$
   \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
$\sum {{n^2} = \dfrac{{n(n + 1)(2n + 1)}}{6}} $
Putting the value of n in this formula, we get
$
   = \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.