If the number of terms in the expansion of {l | vedclass

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(B) The number of terms in the expansion of a multinomial $(a_1 + a_2 + \dots + a_k)^n$ is given by the formula $\binom{n+k-1}{k-1}$.For the expressio

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$729$

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(B) The number of terms in the expansion of a multinomial $(a_1 + a_2 + \dots + a_k)^n$ is given by the formula $\binom{n+k-1}{k-1}$.For the expression $(1 - \frac{2}{x} + \frac{4}{x^2})^n$,we have $k=3$ terms.Thus,the number of terms is $\binom{n+3-1}{3-1} = \binom{n+2}{2} = \frac{(n+2)(n+1)}{2}$.Given $\frac{(n+2)(n+1)}{2} = 28$,we have $(n+2)(n+1) = 56$.Since $8 	imes 7 = 56$,we get $n+2 = 8$,which implies $n = 6$.The sum of the coefficients of an expansion is found by substituting the variable with $1$.Sum of coefficients $= (1 - 2 + 4)^n = (3)^n$.For $n = 6$,the sum is $3^6 = 729$.

If the number of terms in the expansion of ${\left( {1 - \frac{2}{x} + \frac{4}{{{x^2}}}} \right)^n}, x \ne 0$ is $28$,then the sum of the coefficients of all the terms in this expansion is:

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