Coefficient of x50 in (1+x)1000+2x(1+x)999+3x2(1+x)998+⋯+1001x1000 is

S=(1+x)1000+2x(1+x)999+⋯+1001x1000x1+x⋅S=x(1+x)999+2x2(1+x)998+⋯+1001x10011+x1−x1+xS=(1+x)1000+x(1+x)999+x2(1+x)998+⋯+x1000−1001x10011+x⇒11+xS=(1+x)1000+x(1+x)999+⋯+x1000−1001x10011+xS=(1+x)1001+x(1+x)1000+⋯+x1000(1+x)−1001x1001=(1+x)10011−x1+x10011−x1+x−1001x1001=(1+x)1002−x1001(1+x)−1001x1001Coefficient x50=1002C50

Coefficient of x50 in (1+x)1000+2x(1+x)999+3x2(1+x)998+⋯+1001x1000 is

S=(1+x)1000+2x(1+x)999+⋯+1001x1000x1+x⋅S=x(1+x)999+2x2(1+x)998+⋯+1001x10011+x1−x1+xS=(1+x)1000+x(1+x)999+x2(1+x)998+⋯+x1000−1001x10011+x⇒11+xS=(1+x)1000+x(1+x)999+⋯+x1000−1001x10011+xS=(1+x)1001+x(1+x)1000+⋯+x1000(1+x)−1001x1001=(1+x)10011−x1+x10011−x1+x−1001x1001=(1+x)1002−x1001(1+x)−1001x1001Coefficient x50=1002C50

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

Coefficient of $x^{50}$ in $(1 + x)^{1000} + 2 x (1 + x)^{999} + 3 x^{2} (1 + x)^{998} + \dots + 1001 x^{1000}$ is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$^{1002} C_{51}$

$^{1002} C_{50}$

$^{1000} C_{50}$

$^{1001} C_{50}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} S = (1 + x)^{1000} + 2 x (1 + x)^{999} + \dots + 1001 x^{1000} \\ \frac{x}{1 + x} \cdot S = x (1 + x)^{999} + 2 x^{2} (1 + x)^{998} + \dots + \frac{1001 x^{1001}}{1 + x} \\ (1 - \frac{x}{1 + x}) S = (1 + x)^{1000} + x (1 + x)^{999} + x^{2} (1 + x)^{998} + \dots + x^{1000} - \frac{1001 x^{1001}}{1 + x} \\ \Rightarrow \frac{1}{1 + x} S = (1 + x)^{1000} + x (1 + x)^{999} + \dots + x^{1000} - \frac{1001 x^{1001}}{1 + x} \\ S = (1 + x)^{1001} + x (1 + x)^{1000} + \dots + x^{1000} (1 + x) - 1001 x^{1001} \\ = \frac{(1 + x)^{1001} [1 - {(\frac{x}{1 + x})}^{1001}]}{1 - \frac{x}{1 + x}} - 1001 x^{1001} = (1 + x)^{1002} - x^{1001} (1 + x) - 1001 x^{1001} \end{array}$