Sum of infinite series $3 + 5 x + 9 x^{2} + 15 x^{3} + 23 x^{4} + 33 x^{5} + 45 x^{6} + . . .$ is

\frac{2 - 4 x + 3 x^{2}}{{(1 - x)}^{3}}

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\frac{3 - 4 x + 3 x^{2}}{{(1 - x)}^{3}}

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\frac{3 - 4 x - 3 x^{2}}{{(1 - x)}^{3}}

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\frac{3 + 4 x + 3 x^{2}}{{(1 - x)}^{3}}

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Solution

The correct option is B $\frac{3 - 4 x + 3 x^{2}}{{(1 - x)}^{3}}$
Let, $S = 3 + 5 x + 9 x^{2} + 15 x^{3} + 23 x^{4} + 33 x^{5} + 45 x^{6} + . . . \to (1)$
Multiplying (1) by $x$ , we get
$x S = 3 x + 5 x^{2} + 9 x^{3} + 15 x^{4} + 23 x^{5} + 33 x^{6} + 45 x^{7} + . . . \to (2)$

Subtracting (1) & (2), we get
$(1 - x) S = 3 + 2 x + 4 x^{2} + 6 x^{3} + 8 x^{4} + 10 x^{5} + 12 x^{6} + . . . \to (3)$

Multiplying (3) by $x$ , we get
$x (1 - x) S = 3 x + 2 x^{2} + 4 x^{3} + 6 x^{4} + 8 x^{5} + 10 x^{6} + 12 x^{7} + . . . \to (4)$

Subtracting (3) & (4), we get
${(1 - x)}^{2} S = 3 - x + 2 (x^{2} + x^{3} + x^{4} + x^{5} + . . . . .)$
$\Rightarrow {(1 - x)}^{2} S = 3 - x + \frac{2 x^{2}}{1 - x}$ ....[ Sum of Infinite G.P. series ]
$\Rightarrow S = \frac{3 - 4 x + 3 x^{2}}{{(1 - x)}^{3}}$
Ans: B

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$- x^{3} - 4 x^{2} + 9 x - 3$ must be subtracted from $x^{3} - 3 x^{2} + 5 x - 1$ to get $2 x^{3} + x^{2} - 4 x + 2$ . State true or false.

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