Binomial Theorem

if we expand (1+x)ⁿ , then ⁿC₂ represents the coefficient of x²

also ⁿC_r can be written as ⁿC_n-r .

so (¹⁰⁰C₉₈)+(⁹⁹C₉₇)+.........+(³C₁)+(²C₀) =  (¹⁰⁰C₂)+(⁹⁹C₂)+..............+(³C₂)+(²C₂)

so this expression will be equal to coeficient of x² in (1+x)¹⁰⁰ +(1+x)⁹⁹+..............+(1+x)³+(1+x)² 

sum of this series is (1+x)²( (1+x)⁹⁹-1) /  ( (1+x)-1 ) = (1/x) (1+x)²( (1+x)⁹⁹-1 ) = (1/x) ( (1+x)¹⁰¹- (1+x)² )

so we have to find coefficient of x³ in (  (1+x)¹⁰¹- (1+x)² ) . since cofficient of x³ is zero in (1+x)²

so cofficient of x² in this expression is ¹⁰¹C₃