Question

Find the sum to n terms
3,15,35,63...

Answer 1

$\text{The sum to n terms is } \frac{n}{3} [ 4n^2 +6n-1]$

Solution:

The series given is 3, 15, 35, 63 … to n terms

We have to find the sum to n terms

Analyse the given series

$3 = 2^2 - 1\\\\15 = 4^2 - 1\\\\35 = 6^2 - 1\\\\63 = 8^2 - 1$

Thus, we can see a square of increasing even numbers from 2 which are subtracted with 1

Therefore, nth term is given as:

$a_n = (2n)^2-1$

$a_n = 4n^2-1$

Now, we need to find the sum of this series

$S_n = \sum_{n=1}^{n} a_{n}$

$\sum_{n=1}^{n}\left(4 n^{2}-1\right)$

$4 \sum_{n=1}^{n} n^{2}-\sum_{n=1}^{n} 1$

We know that,

Sum of squares of first n natural numbers is:

$\sum_{k=1}^{n} k^{2}=\frac{n(n+1)(2 n+1)}{6}$

Therefore,

$4 \sum_{n=1}^{n} n^{2}-\sum_{n=1}^{n} 1 =4\frac{n(n+1)(2 n+1)}{6} - n$

On solving we get,

$\rightarrow \frac{2n\left(n+1\right)\left(2n+1\right)}{3}-n\\\\\rightarrow \frac{2n\left(n+1\right)\left(2n+1\right) - 3n}{3}\\\\\rightarrow \frac{(2n^2 + 2n)(2n+1) - 3n}{3}\\\\\rightarrow \frac{4n^3 + 2n^2 + 4n^2+2n-3n}{3}\\\\\rightarrow \frac{4n^3+6n^2-n}{3}\\\\\rightarrow \frac{n}{3} [ 4n^2 +6n-1]$

Thus the sum to n terms is found

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