The number of terms in the expansion of (1 + | vedclass

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Question

(C) The given expression is $(1 + x)^{101} (1 - x + x^2)^{100}$.We can rewrite this as $(1 + x) (1 + x)^{100} (1 - x + x^2)^{100}$.$= (1 + x) [(1 + x)

Vedclass · Accepted Answer

$202$

Vedclass · Answer

(C) The given expression is $(1 + x)^{101} (1 - x + x^2)^{100}$.We can rewrite this as $(1 + x) (1 + x)^{100} (1 - x + x^2)^{100}$.$= (1 + x) [(1 + x)(1 - x + x^2)]^{100}$.Using the identity $(a + b)(a^2 - ab + b^2) = a^3 + b^3$,we get $(1 + x)(1 - x + x^2) = 1 + x^3$.So,the expression becomes $(1 + x)(1 + x^3)^{100}$.The expansion of $(1 + x^3)^{100}$ contains $100 + 1 = 101$ terms.Multiplying by $(1 + x)$,we get $(1 + x)(1 + x^3)^{100} = (1 + x^3)^{100} + x(1 + x^3)^{100}$.Each part has $101$ terms,and since the powers of $x$ in $(1 + x^3)^{100}$ are multiples of $3$ $(0, 3, 6, \dots, 300)$ and the powers in $x(1 + x^3)^{100}$ are $1, 4, 7, \dots, 301$,there are no common terms.Therefore,the total number of terms is $101 + 101 = 202$.

The number of terms in the expansion of $(1 + x)^{101} (1 + x^2 - x)^{100}$ in powers of $x$ is

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