જો f(y) = 1 - (y - 1) + {(y - 1)^2} - {(y - 1 | vedclass

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Question

Given function is $f(y)=1-(y-1)+(y-1)^{2}-(y-1)^{3}$

$+\ldots \ldots-(y-1)^{17}$

In the expansion of $(y-1)^{n}$

$T_{r+1}=^{n} C_{r} y^{n-r}(

Vedclass · Accepted Answer

$^{18}{C_3}$

Vedclass · Answer

Given function is $f(y)=1-(y-1)+(y-1)^{2}-(y-1)^{3}$

$+\ldots \ldots-(y-1)^{17}$

In the expansion of $(y-1)^{n}$

$T_{r+1}=^{n} C_{r} y^{n-r}(-1)^{r}$

coeffof $y^{2}$ in $(y-1)^{2}=^{2} \mathrm{C}_{0}$

coeff of $y^{2}$ in $(y-1)^{3}=-3^{3} \mathrm{C}_{1}$

coeffof $y^{2}$ in $(y-1)^{4}=^{4} \mathrm{C}_{2}$

So, coeff of termwise is

$^{2} \mathrm{C}_{0}+^{3} \mathrm{C}_{1}+^{4} \mathrm{C}_{2}+^{5} \mathrm{C}_{3}+\ldots \ldots \ldots+^{17} \mathrm{C}_{15}$

$=1+^{3} C_{1}+^{4} C_{2}+^{5} C_{3}+\ldots \ldots \ldots+^{17} C_{15}$

$=\left(^{3} \mathrm{C}_{0}+^{3} \mathrm{C}_{1}\right)+^{4} \mathrm{C}_{2}+^{5} \mathrm{C}_{3}+\ldots \ldots \ldots+^{17} \mathrm{C}_{15}$

$=^{4} \mathrm{C}_{1}+^{4} \mathrm{C}_{2}+^{5} \mathrm{C}_{3}+\ldots \ldots \ldots+^{17} \mathrm{C}_{15}$

$=^{5} \mathrm{C}_{2}+^{5} \mathrm{C}_{3}+\ldots \ldots \ldots+7^{17} \mathrm{C}_{15}$

$=^{18} \mathrm{C}_{15}=18 \mathrm{C}_{3}$

જો $f(y) = 1 - (y - 1) + {(y - 1)^2} - {(y - 1)^{^3}} + ... - {(y - 1)^{17}},$ હોય તો $y^2$ નો સહગુણક મેળવો.