If $f(y)=1-(y-1)+(y-1)^2-(y-1)^3$ $+\ldots-(y-1)^{17}$ then the coefficient of $y^2$ in it is
If $f(y)=1-(y-1)+(y-1)^2-(y-1)^3$ $+\ldots-(y-1)^{17}$ then the coefficient of $y^2$ in it is
-
${ }^{17} \mathrm{C}_2$
-
${ }^{17} \mathrm{C}_3$
-
${ }^{18} \mathrm{C}_2$
-
${ }^{18} \mathrm{C}_3$
Solution
Given function is
$
\begin{array}{r}
f(y)=1-(y-1)+(y-1)^2-(y-1)^3 \\
+\ldots \ldots \ldots-(y-1)^{17}
\end{array}
$
In the expansion of $(y-1)^n$
$
T_{r+1}={ }^n \mathrm{C}_r y^{n-r}(-1)^r
$
coeff of $y^2$ in $(y-1)^2={ }^2 \mathrm{C}_0$
coeff of $y^2$ in $(y-1)^3={ }^3 \mathrm{C}_1$
coeff of $y^2$ in $(y-1)^4={ }^4 \mathrm{C}_2$
So, coeff of termwise is
$
\begin{aligned}
& { }^2 \mathrm{C}_0+{ }^3 \mathrm{C}_1+{ }^4 \mathrm{C}_2+{ }^5 \mathrm{C}_3+\ldots \ldots \ldots+{ }^{17} \mathrm{C}_{15} \\
& =1+{ }^3 \mathrm{C}_1+{ }^4 \mathrm{C}_2+{ }^5 \mathrm{C}_3+\ldots \ldots \ldots . .+{ }^{17} \mathrm{C}_{15} \\
& =\left({ }^3 \mathrm{C}_0+{ }^3 \mathrm{C}_1\right)+{ }^4 \mathrm{C}_2+{ }^5 \mathrm{C}_3+\ldots \ldots \ldots . .+{ }^{17} \mathrm{C}_{15} \\
& ={ }^4 \mathrm{C}_1+{ }^4 \mathrm{C}_2+{ }^5 \mathrm{C}_3+\ldots \ldots \ldots+{ }^{17} \mathrm{C}_{15}
\end{aligned}
$
$
\begin{aligned}
& ={ }^5 \mathrm{C}_2+{ }^5 \mathrm{C}_3+\ldots \ldots \ldots+{ }^{17} \mathrm{C}_{15} \\
& ={ }^{18} \mathrm{C}_{15}={ }^{18} \mathrm{C}_3
\end{aligned}
$
Asked in: JEE Main 2012 (07 May Online)
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