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

Last updated date: 24th Sep 2023
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