Find the coefficient of \( x^{100} \) in;-1 \[ (1+x)+2(1+x)^{2}+3(1+x)^{3}+\ldots 1000(1+x)^{1000} \]

(1+x) + 2(1+x)² + 3(1+x)³ + .....+ 1000 (1+x)¹⁰⁰⁰

Now,

1 + (1+x) + (1+x)² + (1+x)³ + .....+ 1+x)¹⁰⁰⁰ = \(\frac{1-(1+x)^{1001}}{1-(1+x)}\) 

(∵1+x+x²+......+xⁿ = \(\frac{1-(1+x)^{n+1}}{1-x}\))

By differentiating both sides w.r.t.x, we get

⇒ 1+2(1+x)+3(1+x)² + .....+1000(1+x)⁹⁹⁹ = \(\frac{-[x\times -1001(1+x)^{1000}-(1-(1+x)^{1001}\times 1)]}{x^2}\)

⇒ 1+2(1+x)+3(1+x)² + .....+1000(1+x)⁹⁹⁹ = \(\frac{1001x(1+x)^{1000}+1-(1+x)^{1001}}{x^2}\)

Multiplying both sides by (1+x), we get

(1+x) + 2(1+x)² + 3(1+x)³ +....+1000(1+x)1000 = \(\frac{1001x(1+x)^{1001}-(1+x)^{1002}}{x^2}\)

= \(\frac{1001(1+x)^{1001}}{x}\) - \(\frac{(1+x)^{1002}}{x^2}\) + \(\frac{1+x}{x^2}\)

∴ coefficient of x¹⁰⁰ in ((1+x)+2(1+x)²+.....+1000(1+x)¹⁰⁰⁰)

= 1001 x coefficient of x¹⁰¹ in (1+x)¹⁰⁰¹ - coefficient of x¹⁰² in (1+x)¹⁰⁰².

= 1001 x \(^{1001}C_{101}\) - \(^{1002}C_{102}\)

= 1001 x \(\frac{(1001)!}{(101)!(900)!}\) - \(\frac{(1002)!}{(102)!(900)!}\)

= 1001 x \(\frac{(1001)!}{(101)!(900)!}\) - \(\frac{1002\times (1001)!}{102\times(101)!(900)!}\)

= \(\frac{(1001)!}{(101)!(900)!}\) (1001 - \(\frac{1002}{102}\))

= \(^{1001}C_{101}\)(1001 - \(\frac{167}{17}\))

= \(^{1001}C_{101}\) x \(\frac{16850}{17}\) 

Hence,

The coefficients of x¹⁰⁰ in [(1+x) + 2(1+x)²+3(1+x)³ + ....+1000(1+x)¹⁰⁰⁰] is \(\frac{16850}{17}\) x \(^{1001}C_{101}\).