JEE MAIN - Mathematics (2016 - 9th April Morning Slot - No. 1)
For x $$ \in $$ R, x $$ \ne $$ -1,
if (1 + x)2016 + x(1 + x)2015 + x2(1 + x)2014 + . . . . + x2016 =
$$\sum\limits_{i = 0}^{2016} {{a_i}} \,{x^i},\,\,$$ then a17 is equal to :
if (1 + x)2016 + x(1 + x)2015 + x2(1 + x)2014 + . . . . + x2016 =
$$\sum\limits_{i = 0}^{2016} {{a_i}} \,{x^i},\,\,$$ then a17 is equal to :
$${{2017!} \over {17!\,\,\,2000!}}$$
$${{2016!} \over {17!\,\,\,1999!}}$$
$${{2017!} \over {2000!}}$$
$${{2016!} \over {16!}}$$
Explanation
Assume,
P = (1 + x)2016 + x(1 + x)2015 + . . . . .+ x2015 . (1 + x) + x2016 . . . . .(1)
Multiply this with $$\left( {{x \over {1 + x}}} \right),$$
$$\left( {{x \over {1 + x}}} \right)P = $$ x(1 + x)2015 + x2(1 + x)2014 +
. . . . . . + x2016 + $${{{x^{2017}}} \over {1 + x}}$$ . . . . . (2)
Performing (1) $$-$$ (2), we get
$${P \over {1 + x}} = $$ (1 + x)2016 $$-$$ $${{{x^{2017}}} \over {1 + x}}$$
$$ \Rightarrow $$ P = (1 + x)2017 $$-$$ x2017
$$ \therefore $$ a17 = coefficient of x17 $$=$$ 2017C17 $$=$$ $${{2017!} \over {17!\,\,2000!}}$$
P = (1 + x)2016 + x(1 + x)2015 + . . . . .+ x2015 . (1 + x) + x2016 . . . . .(1)
Multiply this with $$\left( {{x \over {1 + x}}} \right),$$
$$\left( {{x \over {1 + x}}} \right)P = $$ x(1 + x)2015 + x2(1 + x)2014 +
. . . . . . + x2016 + $${{{x^{2017}}} \over {1 + x}}$$ . . . . . (2)
Performing (1) $$-$$ (2), we get
$${P \over {1 + x}} = $$ (1 + x)2016 $$-$$ $${{{x^{2017}}} \over {1 + x}}$$
$$ \Rightarrow $$ P = (1 + x)2017 $$-$$ x2017
$$ \therefore $$ a17 = coefficient of x17 $$=$$ 2017C17 $$=$$ $${{2017!} \over {17!\,\,2000!}}$$
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