Sum of $n$ terms of the series $\sqrt {2}+\sqrt {8}+\sqrt {18}+\sqrt {32}+.......$ is
- $\dfrac {n(n+1)}{2}$
- $\dfrac {n(n+1)}{\sqrt {2}}$
- $2n(n+1)$
- $1$
Series is sqrt(2), 2*sqrt(2), 3*sqrt(2), 4*sqrt(2)... This is an AP with a = sqrt(2) and d = sqrt(2). Sum of n terms = n/2 * (2a + (n-1)d) = n/2 * (2*sqrt(2) + (n-1)sqrt(2)) = n/2 * (sqrt(2) * (n+1)) = n(n+1)/sqrt(2).
The given series can be rewritten by simplifying the square roots: sqrt(2) + sqrt(8) + sqrt(18) + sqrt(32) becomes sqrt(2) + 2sqrt(2) + 3sqrt(2) + 4sqrt(2). This forms an arithmetic progression with the first term a = sqrt(2) and common difference d = sqrt(2). Using the sum formula for n terms of an arithmetic progression, S = n/2 [2a + (n-1)d], we substitute the values to get S = n/2 [2sqrt(2) + (n-1)sqrt(2)], which simplifies to n/2 [sqrt(2)(n+1)]. This further simplifies to n(n+1) / sqrt(2).