Find the coefficient of x50 in the expansion : (1+x)1000 + 2x(1+x)999 +3x2(1+x)998+ - Maths - Binomial Theorem

Question

Find the coefficient of x50 in the expansion :

(1+x)1000 + 2x(1+x)999
+3x2(1+x)998+…………………
+1001x1000

Utsav · Accepted Answer

Let S=(1+x)1000+2x(1+x)99+3x2(1+x)998+ +1000(1+x)x999+1001x1000
(1)⇒xS1+x=x(1+x)999+2x2(1+x)998+ 1000x1000+1001x10011+x
(2)Now (1)-(2) ,we getS-xS1+x=(1+x)1000+x(1+x)999+x2(1+x)998+
x1000-1001x10011+x⇒S1+x=(1+x)10001-(x1+x)10011-x1+x-1001x10011+x⇒S1+x=(1+x)1001-x1001-1001x10011+x
⇒S=(1+x)1002-x1001(1+x)-1001x1001⇒S=(1+x)1002-1002x1001-x1002Now
coefficient of x50in S = coefficient of x50in (1+x)1002=1002C50

Find the coefficient of x50 in the expansion : (1+x)1000 + 2x(1+x)999 +3x2(1+x)998+…………………..+1001x1000

Find the coefficient of x⁵⁰ in the expansion :

(1+x)¹⁰⁰⁰ + 2x(1+x)⁹⁹⁹ +3x²(1+x)⁹⁹⁸+…………………..+1001x¹⁰⁰⁰