If S(x) = (1+x) + 2(1+x)^2 + 3(1+x)^3 + ldots | vedclass

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(C) Let $y = 1+x$. Then $S(x) = y + 2y^2 + 3y^3 + \ldots + 60y^{60}$.
This is an arithmetico-geometric series.
$yS = y^2 + 2y^3 + \ldots + 59y^{60}

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(C) Let $y = 1+x$. Then $S(x) = y + 2y^2 + 3y^3 + \ldots + 60y^{60}$.
This is an arithmetico-geometric series.
$yS = y^2 + 2y^3 + \ldots + 59y^{60} + 60y^{61}$.
Subtracting the two equations: $(1-y)S = y + y^2 + y^3 + \ldots + y^{60} - 60y^{61}$.
Since $y = 1+x$,$1-y = -x$.
$-xS = \frac{y(y^{60}-1)}{y-1} - 60y^{61} = \frac{(1+x)((1+x)^{60}-1)}{x} - 60(1+x)^{61}$.
For $x=60$,$y=61$:
$-60S(60) = \frac{61(61^{60}-1)}{60} - 60(61)^{61}$.
Multiply by $-60$:
$3600 S(60) = (60)^2 S(60) = 60(61)^{61} - 61(61^{60}-1) = 60(61)^{61} - 61^{61} + 61 = 59(61)^{61} + 61$.
Comparing with $a(b)^b + b$,we get $a=59$ and $b=61$.
Thus,$a+b = 59+61 = 120$.

If $S(x) = (1+x) + 2(1+x)^2 + 3(1+x)^3 + \ldots + 60(1+x)^{60}$,$x \neq 0$,and $(60)^2 S(60) = a(b)^b + b$ where $a, b \in N$,then $(a+b)$ is equal to:

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