If f(y) = 1 - (y - 1) + (y - 1)^2 - (y - 1)^3 | vedclass

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(D) The given expression is a finite geometric series with first term $a = 1$, common ratio $r = -(y - 1)$, and $n = 18$ terms.
The sum is given by $

Vedclass · Accepted Answer

$^{18}C_3$

Vedclass · Answer

(D) The given expression is a finite geometric series with first term $a = 1$, common ratio $r = -(y - 1)$, and $n = 18$ terms.
The sum is given by $f(y) = \frac{1 - (-(y - 1))^{18}}{1 - (-(y - 1))} = \frac{1 - (y - 1)^{18}}{y}$.
Thus, $f(y) = \frac{1}{y} - \frac{(y - 1)^{18}}{y}$.
Expanding $(y - 1)^{18}$ using the binomial theorem: $(y - 1)^{18} = \sum_{k=0}^{18} {^{18}C_k} y^k (-1)^{18-k}$.
Therefore, $\frac{(y - 1)^{18}}{y} = \sum_{k=0}^{18} {^{18}C_k} y^{k-1} (-1)^{18-k}$.
To find the coefficient of $y^2$, we set $k - 1 = 2$, which gives $k = 3$.
The term for $k = 3$ is ${^{18}C_3} y^2 (-1)^{18-3} = -{^{18}C_3} y^2$.
Since $f(y) = \frac{1}{y} - \sum_{k=0}^{18} {^{18}C_k} y^{k-1} (-1)^{18-k}$, the coefficient of $y^2$ is $-(-{^{18}C_3}) = {^{18}C_3}$.

If $f(y) = 1 - (y - 1) + (y - 1)^2 - (y - 1)^3 + \dots - (y - 1)^{17}$,then the coefficient of $y^2$ in it is

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