The number of terms in ( 1 +x)^101(1+x^2-x)^100
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(1+x)
101
(1−x+x
2
)
100
can be written as
=(1+x)(1+x)
100
(1−x+x
2
)
100
=(1+x)[(1+x)(1−x+x
2
)]
100
=(1+x)(1+x
3
)
100
=(1+x
3
)
100
+x(1+x
3
)
100
Number of terms in (1+x
3
)
100
=101[terms are a
0
+a
1
x
3
+a
2
x
6
+............+a
1
00x
300
]
Number of terms in x(1+x
3
)
100
=101[terms are a
0
x+a
1
x
4
+a
2
x
7
+............+a
1
00x
301
]
since there are no common terms
.
So, number of terms in (1+x)(1+x
3
)
100
=101+101=202
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