Find the number of distinct terms in the expansion of
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First of all we write its general term :(Using Multinomial)
r 1 ! r ! 2 r 3 ! 1 0 0 ! ⋅ ( x 3 ) r 1 ( 1 ) r 2 ( x 3 1 ) r 3
where r 1 , r 2 , r 3 are integers and r 1 + r 2 + r 3 = 1 0 0 . On simplification ;
r 1 ! r ! 2 r 3 ! 1 0 0 ! ⋅ ( x ) 3 ( r 1 − r 3 ) .
Now,since r 1 and r 3 are integers ( r 1 − r 3 ) will also be an integer (say r ) where r ∈ [ − 1 0 0 , 1 0 0 ] Now distinct terms means terms having different powers of x because similar powers of x will add up together .So we will have powers of x ranging from x − 3 0 0 , x − 2 9 7 , x − 2 9 4 . . to . . x − 3 , x 0 , x 3 . . to . . x 2 9 4 , x 2 9 7 , x 3 0 0 which are 2 0 1 terms.