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(B) The expansion of $(1 + x)^{100}$ has $101$ terms (powers $x^0$ to $x^{100}$).
The expansion of $(1 + x^2)^{100}$ has $101$ terms (powers $x^0, x^

Vedclass · Accepted Answer

$218$

Vedclass · Answer

(B) The expansion of $(1 + x)^{100}$ has $101$ terms (powers $x^0$ to $x^{100}$).
The expansion of $(1 + x^2)^{100}$ has $101$ terms (powers $x^0, x^2, \dots, x^{200}$).
The expansion of $(1 + x^3)^{100}$ has $101$ terms (powers $x^0, x^3, \dots, x^{300}$).
Let $S_1, S_2, S_3$ be the sets of powers of $x$ in each expansion.
$|S_1| = 101, |S_2| = 101, |S_3| = 101$.
Using the Principle of Inclusion-Exclusion,the number of distinct terms is $|S_1 \cup S_2 \cup S_3| = |S_1| + |S_2| + |S_3| - (|S_1 \cap S_2| + |S_2 \cap S_3| + |S_1 \cap S_3|) + |S_1 \cap S_2 \cap S_3|$.
$|S_1 \cap S_2|$ contains powers $x^k$ where $k$ is a multiple of $2$ and $0 \le k \le 100$,so $51$ terms.
$|S_1 \cap S_3|$ contains powers $x^k$ where $k$ is a multiple of $3$ and $0 \le k \le 100$,so $34$ terms.
$|S_2 \cap S_3|$ contains powers $x^k$ where $k$ is a multiple of $6$ and $0 \le k \le 100$,so $17$ terms.
$|S_1 \cap S_2 \cap S_3|$ contains powers $x^k$ where $k$ is a multiple of $6$ and $0 \le k \le 100$,so $17$ terms.
Total terms $= 101 + 101 + 101 - (51 + 34 + 17) + 17 = 303 - 102 + 17 = 218$.

The total number of terms in the expansion of $[(1 + x)^{100} + (1 + x^2)^{100} + (1 + x^3)^{100}]$ is -

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