Question

Sum of n terms of the series  `sqrt2+sqrt8+sqrt18+sqrt32+....` is ______.

विकल्प

• `(n(n+2))/sqrt2`
• `sqrt2  n(n+1)`

• `(n(n+1))/sqrt2`

• 1

Answer 1

Sum of n terms of the series  `sqrt2+sqrt8+sqrt18+sqrt32+....` is `underline((n(n+1))/sqrt2)`.

Explanation:-

In the given problem, we need to find the sum of terms for a given arithmetic progression,

`sqrt(2) , sqrt(8), sqrt(18) , sqrt(32) , .....`

So, here we use the following formula for the sum of n terms of an A.P.,

`S_n = n/2 [2a + (n-1)d]`

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

Here,

Common difference of the A.P. (d) =  a₂ - a₁

`= sqrt(8) - sqrt(2)`

`=2sqrt(2) - sqrt(2)`

`= sqrt(2)`

Number of terms (n) = n

First term for the given A.P. (a) = `sqrt(2) `

So, using the formula we get,

`S_n = n/2 [2(sqrt(2)) + (n-1) (sqrt(2))]`

      ` = (n/2) [2sqrt(2) + ( sqrt(2n) - sqrt(2) ]` 

     `= (n/2) (2sqrt(2) + sqrt(2n) - sqrt(2))`

     `= (n/2)(sqrt(2) + sqrt(2n)  ) `

Now, taking `sqrt(2)`  common from both the terms inside the bracket we get,

`= (n/2) sqrt(2) (n + 1)`

`=(n(n + 1))/(sqrt(2)`

Therefore, the sum of first n terms for the given A.P. is

`S_n = (n(n + 1))/sqrt(2)`.