JEE Mains · Maths · STD 11 - 7. binomial theoram
For \(x\, \in \,R\,,\,x\, \ne \, - 1,\) if \({(1 + x)^{2016}} + x{(1 + x)^{2015}} + {x^2}{(1 + x)^{2014}} + ....{x^{2016}} = \sum\limits_{i = 0}^{2016} {{a_i\,}{\,x^i}} ,\) then \(a_{17}\) is equal to
- A \(\frac{{2017\,!\,}}{{17\,!\,2000\,!}}\)
- B \(\frac{{2016\,!\,}}{{17\,!\,1999\,!}}\)
- C \(\frac{{2016\,!\,}}{{16\,!}}\)
- D \(\frac{{2017\,!\,}}{{2000\,!}}\)
Answer & Solution
Correct Answer
(A) \(\frac{{2017\,!\,}}{{17\,!\,2000\,!}}\)
Step-by-step Solution
Detailed explanation
\(S=(1+x)^{2016}+x(1+x)^{2015}+x^{2}(1+x)^{2014}\) \(+\ldots+x^{2015}(1+x)+x^{2016}........(i)\) \(\left(\frac{x}{1+x}\right) S=x(1+x)^{2015}+x^{2}(1+x)^{2014}\) \(+\ldots +x^{2016}+\frac{x^{2017}}{1+x}........(ii)\) Subtracting \((i)\) from \((ii)\)…
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