JEE Mains · Maths · STD 11 - 8. sequence and series
The coefficient of \(x^{48}\) in \((1+x)+2(1+x)^{2}+3(1+x)^{3}+....+100(1+x)^{100}\) is equal to:
- A \(100 \cdot ^{100}C_{49} - ^{100}C_{50}\)
- B \(^{100}C_{50} + ^{101}C_{49}\)
- C \(100 \cdot^{100}C_{49} - ^{106}C_{48}\)
- D \(100 \cdot ^{101}C_{49} - ^{101}C_{50}\)
Answer & Solution
Correct Answer
(D) \(100 \cdot ^{101}C_{49} - ^{101}C_{50}\)
Step-by-step Solution
Detailed explanation
Let \(1+ x = r\) \(\therefore S=1r+2r^2+3r^3+\ldots \ldots+100 r^{100}\ldots \ldots\)(1)(AGP) \(rS =1r ^2+2r ^3+\ldots \ldots+99 r ^{100}+100 r ^{101} \ldots \ldots\).(2) (1) - (2) gives \(S=-\frac{(1+x)^{101}}{x^2}+\frac{1}{x^2}+\frac{100(1+x)^{101}}{x}\) ∴ coefficient…
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