IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

deep01

Find the coefficient of x⁵⁰ in the polynomial

(1+x)+2(1+x)²+3(1+x)³+________________+1000(1+x)¹⁰⁰⁰??

pls also explain ur answer.

Srimann

Let S=(1+x)+2(1+x)^2+...+1000(1+x)^1000
(1+x)S=(1+x)^2+2(1+x)^3+...+999(1+x)^1000+1000(1+x)^1001
subtract,
-xS=(1+x)+(1+x)^2+...+(1+x)^1000-1000(1+x)^1001
=

deep01

now what next we have to do???

yahiyafirdous

Let y= 1+ x

S_n= y + 2y²+ 3y³+.................+(n-1)y^n-1 + nyⁿ ...................(1)

yS_n= y² +2y³+ +(n-2)y^n-1 + (n-1)yⁿ + ny^{n+1..........}(2)

Subtracting (2) from (1) we get:

(1-y)S_n = y + y² +y³ +..................................+yⁿ - nyⁿ⁺¹ where n=1000

(1-y)S_n = y(1-yⁿ)/(1-y) - nyⁿ⁺¹

S_n =y(1-yⁿ)/(1-y)² - nyⁿ⁺¹/(1-y)

Substitute y=1+x

S_n=[(1+x) - (1+x)ⁿ⁺¹]/x² + n(1+x)ⁿ⁺¹/x

=(1+x)/x² - x^-2(1+x)¹⁰⁰¹ + 1000x^-1 (1+x)¹⁰⁰¹

Now required Co-eff= -¹⁰⁰¹C₅₂+ 1000¹⁰⁰¹C₅₁ Ans

Ask & Discuss Questions with Community & Experts