Binomial Theorem Question - None

Question

The coefficient of x⁵⁰ in the expansion of S = (1 + x)¹⁰⁰⁰ + 2x(1 + x)⁹⁹⁹ + 3x²(1 + x)⁹⁹⁸ + ... + 10001x¹⁰⁰⁰is

Accepted Answer

S = (1 + x)¹⁰⁰⁰ + 2x(1 + x)⁹⁹⁹ + 3x²(1 + x)⁹⁹⁸ + ... + 1000(1 + x) x⁹⁹⁹ + 1001 x¹⁰⁰⁰ ...(i)
$\Rightarrow \frac{x S}{1+x}$ = x(1 + x)⁹⁹⁹ + 2x²(1 + x)⁹⁹⁸ + ... + 1000 x¹⁰⁰⁰ + 100$\frac{x^{1001}}{1+x}$ ...(ii)
(i) and (ii) give
S - $\frac{x \mathrm{~S}}{1+x}$
= {(1 + x)¹⁰⁰⁰ + x(1 + x)⁹⁹⁹ + x²(1 + x)⁹⁹⁸ + ... + x¹⁰⁰⁰} - 1001$\frac{x^{1001}}{1+x}$
L.H.S. = $\frac{\mathrm{S}}{1+x}$
and R.H.S.
= (1 + x)¹⁰⁰⁰ $\left\{\frac{1-\left(\frac{x}{1+x}\right)^{1001}}{1-\frac{x}{1+x}}\right\}$ - 1001$\frac{x^{1001}}{1+x}$
$\Rightarrow \frac{S}{1+x}$ = (1 + x)¹⁰⁰¹ - x¹⁰⁰¹ - 1001$\frac{x^{1001}}{1+x}$
$\Rightarrow$ S = (1 + x)¹⁰⁰² - x¹⁰⁰¹ (1 + x) - 1001 x¹⁰⁰¹
$\Rightarrow$ S = (1 + x)¹⁰⁰² - 1002 x¹⁰⁰¹ - x¹⁰⁰²
The coefficient of x⁵⁰ in S
= coefficient of x⁵⁰ in (1 + x)¹⁰⁰²
= ¹⁰⁰²C₅₀

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