Question

Please solve this need urgent

Answer 1

$\underline{\underline{\mathfrak{\green{Answer:-}}}}$

(C) $Sn = \dfrac{n(n+1)}{\sqrt{2}}$

$\underline{\underline{\mathfrak{\green{Explanation:-}}}}$

Given:

The series is-

$\sqrt{2} + \sqrt{8}+\sqrt{18} +\sqrt{32}+$........

To Find:

Sum of "n" terms of the series

Solution:

The Given series is-

$\sqrt{2} + \sqrt{8}+\sqrt{18} +\sqrt{32}+$........

Here,

$a \:\:or\:\: a1= \sqrt{2}$

$a2= \sqrt{8} = 2 \sqrt{2}$

$a3 = \sqrt{18} = 3 \sqrt{2}$

$d = a2 - a1 = a3- a2$

$d = \sqrt{8} - \sqrt{2} = \sqrt{18}- \sqrt{8}$

$d = 2\sqrt{2} - \sqrt{2} =3 \sqrt{2}- 2\sqrt{2}$

$d = \sqrt{2} = \sqrt{2}$

Since,

Common difference is equal

Therefore, the above series is in AP

By the sum of "n" terms formula-

☆$\boxed{\pink{Sn = \dfrac{n}{2}[2a+(n-1)d]}}$

By sub. the values-

$\mathsf{Sn = \dfrac{n}{2}[2(\sqrt{2)}+(n-1)(\sqrt{2})]}$

$\mathsf{Sn = \dfrac{n}{2}[2\sqrt{2}+\sqrt{2}n-\sqrt{2}]}$

$\mathsf{Sn = \dfrac{n}{2}[\sqrt{2}+\sqrt{2}n]}$

$\mathsf{Sn = \dfrac{\sqrt{2}n(n+1)}{2}}$

$\mathsf{Sn = \dfrac{\sqrt{2}n(n+1)}{\sqrt{2} \times \sqrt{2}}}$

$\mathsf{Sn = \dfrac{\cancel{\sqrt{2}}n(n+1)}{\cancel{\sqrt{2}} \times \sqrt{2}}}$

$\mathsf{Sn = \dfrac{n(n+1)}{\sqrt{2}}}$

Hence,

Sum of n terms is

$\dfrac{n(n+1)}{\sqrt{2}}$

Answer 2

Solution is in attachment

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