Question

Find the sum to n terms of series: 5+11+19+29+41.............It is G.P of 11th Std not A.P.

Answer 1

n(n+1)(2n+1/3+1)-1
n(n+1)(2n+4)/6-1

Answer 2

Answer:

\frac{n(n+1)(2n+4)}{6}-1

Step-by-step explanation:

In the given series,

We have the series,

5 + 11 + 19 + 29 + 41 + .................

So,

We can simply write it as,

5 + 11 + 19 + 29 + 41 + ................. = (2² + 1) + (3² + 2) + (4² + 3) + (5² + 4) + ............

On rearranging, we get,

= (2² + 3² + 4² + 5² +.......) + (1 + 2 + 3 + ........)

Now, we know that sum of squares of first 'n' terms is given by,

$Sum=\frac{n(n+1)(2n+1)}{6}$

So,

We can write the first part as,

$2^{2} + 3^{2} + 4^{2} + 5^{2} +.......=\frac{n(n+1)(2n+1)}{6}-1$

also,

Sum of the second half is given by,

$Sum=\frac{n}{2}(2a+(n-1)d)\\Sum=\frac{n}{2}(2+(n-1))=\frac{n}{2}(n+1)$

Therefore, the total sum of the 'n' terms is given by,

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

Therefore, the sum of the the terms is given by,

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