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(B) The general term in the expansion of $(x^{n} + 2x^{-5})^{7}$ is given by $T_{r+1} = {}^{7}C_{r} (x^{n})^{7-r} (2x^{-5})^{r} = {}^{7}C_{r} \cdot 2^

Vedclass · Accepted Answer

$57$

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(B) The general term in the expansion of $(x^{n} + 2x^{-5})^{7}$ is given by $T_{r+1} = {}^{7}C_{r} (x^{n})^{7-r} (2x^{-5})^{r} = {}^{7}C_{r} \cdot 2^{r} \cdot x^{n(7-r) - 5r}$.The coefficients are $C_{r} = {}^{7}C_{r} \cdot 2^{r}$ for $r = 0, 1, 2, 3, 4, 5, 6, 7$.We are given the sum of coefficients of positive powers of $x$ is $939$.Let's calculate the coefficients:$r=0$: $C_{0} = {}^{7}C_{0} \cdot 2^{0} = 1 \cdot 1 = 1$Power: $7n$$r=1$: $C_{1} = {}^{7}C_{1} \cdot 2^{1} = 7 \cdot 2 = 14$Power: $6n-5$$r=2$: $C_{2} = {}^{7}C_{2} \cdot 2^{2} = 21 \cdot 4 = 84$Power: $5n-10$$r=3$: $C_{3} = {}^{7}C_{3} \cdot 2^{3} = 35 \cdot 8 = 280$Power: $4n-15$$r=4$: $C_{4} = {}^{7}C_{4} \cdot 2^{4} = 35 \cdot 16 = 560$Power: $3n-20$$r=5$: $C_{5} = {}^{7}C_{5} \cdot 2^{5} = 21 \cdot 32 = 672$Power: $2n-25$Summing the coefficients: $1 + 14 + 84 + 280 + 560 = 939$.This corresponds to the terms where the power of $x$ is positive,i.e.,$r=0, 1, 2, 3, 4$.Thus,the power of $x$ for $r=4$ must be $\geq 0$ and for $r=5$ must be $$3n - 20 \geq 0 \implies n \geq \frac{20}{3} \approx 6.66$.$2n - 25 Since $n$ is an integer,$n \in \{7, 8, 9, 10, 11, 12\}$.The sum of these values is $7 + 8 + 9 + 10 + 11 + 12 = 57$.

If the sum of the coefficients of all the positive powers of $x$ in the binomial expansion of $(x^{n} + \frac{2}{x^{5}})^{7}$ is $939$,then the sum of all the possible integral values of $n$ is

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