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Q.
Infinite number of triangles are formed as shown in figure. If total area of these triangles is A then $8 A$ is equal to
Sequences and Series
Solution:
We have,
$A =\left(\frac{1}{2} \times \frac{1}{3} \times 1\right)+\left(\frac{1}{2}+\frac{1}{9} \times 2\right)+\left(\frac{1}{2} \times \frac{1}{27}+3\right)+\ldots \infty$
$=\underbrace{\frac{1}{2}\left(\frac{1}{3}+\frac{2}{9}+\frac{3}{27}+\ldots \infty\right)}_{ s }=\frac{ S }{2}$
Let $S=\frac{1}{3}+\frac{2}{3^{2}}+\frac{3}{3^{3}}+\ldots \infty$, then
$\frac{ S }{3}=0+\frac{1}{3^{2}}+\frac{2}{3^{3}}+\frac{3}{3^{4}}+\ldots \infty$
(Subtracting)
$\Rightarrow \frac{2 S }{3}=\frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+\ldots \infty$
$=\frac{\frac{1}{3}}{1-\frac{1}{3}}=\frac{\frac{1}{3}}{\frac{2}{3}}=\frac{1}{2} $
$\Rightarrow S =\frac{3}{4}$
$\therefore A =\frac{ S }{2}=\frac{3}{8} $
$\Rightarrow 8 A =3$
