Find the number of terms in the expansion in $$(1+x)^{100}+(1+x^2)^{100}+ (1+x^3)^{100}$$
Note that $(1+x)^{100}$ gives 101 terms ranging from x^0 to x^100, since all the terms are positive we have at least 101 terms. $(1+x^2)^{100}$ and $(1+x^3)^{100}$ further adds the powers of x which are between 100 and 201, the first one adds the even powers and the second one adds all powers which are divisble by 3. Hence, the total such terms is given by adding all the multiples of 2 and multiple of 3's and subtracting the multiples of 6 by PIE.
So the final result comes as $101+50+33-17=166$
However this seems to be incorrect answer, I'm not able to figure out what I have done incorrectly.
Please help. Thanks.
Each addend gives you $101$ terms, but there are $50$ terms with even exponent $<101$ in the expansion of the second addend. There are also $34$ terms with exponent $<101$ in the expansion of the third addend and $17$ terms with multiple of $6$ exponent from $102$ to $198$.
That makes $$303-50-34-17=202$$
Another way:
The number of terms is $101$ that is the numbers form $0$ to $100$ plus $50$ even numbers from $102$ to $200$ plus $17$ odd, multiple of $3$ numbers from $105$ to $195$, plus $34$ multiple of $3$ numbers from $201$ to $300$.
That makes $$101+50+17+34=202$$