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प्रश्न
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
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उत्तर
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)`.
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