The
nth term = $\dfrac{3n+1}{3^{n+1}}$ n is from 0 to infinite
= $\dfrac{n}{3^n}+ \dfrac{1}{3}\cdot\dfrac{1}{3^n}$
The second term is GP and the sum upto infinite terms is $\dfrac{1}{3}\cdot\dfrac{1}{1-\frac{1}{3}} = \dfrac{1}{2}$
let $f(x) = 1 + x + x^2 + \cdots = \dfrac{1}{1-x}$
then differentiate both sides wrt x to get
$f'(x) = 1 + 2x + 3x^2 + \cdots. = \dfrac{1}{(1-x)^2}$
multiply by x to get
$xf'(x) = x + 2x^2 + 3x^3 \cdots = \dfrac{x}{(1-x)^2}$
$x = \dfrac{1}{3}$ gives the 1st sum as $\dfrac{\frac{1}{3}}{(\frac{2}{3})^2}= \dfrac{9}{4}\cdot\dfrac{1}{3} = \dfrac{3}{4}$
so sum = $\dfrac{3}{4}+ \dfrac{1}{2} = \dfrac{5}{4}$
= $\dfrac{n}{3^n}+ \dfrac{1}{3}\cdot\dfrac{1}{3^n}$
The second term is GP and the sum upto infinite terms is $\dfrac{1}{3}\cdot\dfrac{1}{1-\frac{1}{3}} = \dfrac{1}{2}$
let $f(x) = 1 + x + x^2 + \cdots = \dfrac{1}{1-x}$
then differentiate both sides wrt x to get
$f'(x) = 1 + 2x + 3x^2 + \cdots. = \dfrac{1}{(1-x)^2}$
multiply by x to get
$xf'(x) = x + 2x^2 + 3x^3 \cdots = \dfrac{x}{(1-x)^2}$
$x = \dfrac{1}{3}$ gives the 1st sum as $\dfrac{\frac{1}{3}}{(\frac{2}{3})^2}= \dfrac{9}{4}\cdot\dfrac{1}{3} = \dfrac{3}{4}$
so sum = $\dfrac{3}{4}+ \dfrac{1}{2} = \dfrac{5}{4}$
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