JEE Mains · Maths · STD 11 - 7. binomial theoram
The sum of the coefficients of \(x^{499}\) and \(x^{500}\) in \((1+x)^{1000}+x(1+x)^{999}+x^{2}(1+x)^{998}+.......+x^{1000}\) is
- A \({}^{1001}C_{501}\)
- B \({}^{1002}C_{500}\)
- C \({}^{1002}C_{501}\)
- D \({}^{1000}C_{501}\)
Answer & Solution
Correct Answer
(B) \({}^{1002}C_{500}\)
Step-by-step Solution
Detailed explanation
\(S=(1+x)^{1000}+x(1+x)^{999}+x^{2}(1+x)^{998}+….+x^{1000}\) \(=(1+x)^{1001}\frac{(1-(\frac{x}{1+x})^{1001})}{1-\frac{x}{1+x}}\) \(=(1+x)^{1001}-x^{1001}\) Required sum=\({}^{1001}C_{499}+{}^{1001}C_{500}={}^{1002}C_{500}\)
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