For x [ ] R, x [ ] -1, if (1 + x)2016 + x(1 + x)2015 + x2(1 + x)2014 + .

Question

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 :

Accepted Answer

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 $=$ $${{2017!} \over {17!\,\,2000!}}$$

Solutions · 1