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If the sum of the coefficient of all the positive powers of \mathrm{x}, in the Binomial expansion of \mathrm{\left(x^{n}+\frac{2}{x^{5}}\right)^{7}} is \mathrm{939}, then the sum of all the possible integral values of \mathrm{n} is ________.

Option: 1

57


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\mathrm{ \left(a^{n}+\frac{2}{x^{5}}\right)^{7}={ }^{7} c_{0} x^{7 n}+{ }^{7} c_{1} \cdot x^{6 n-5} \cdot 2+{ }^{7} c_{2} \cdot x^{5 n-10} \cdot 2^{2}+{ }^{7} / 3 \cdot x^{4 n-15} \cdot 2^{3} }\\

             \mathrm{+{ }^{7} c_{4} x^{3 n-20} 2^{4}+{ }^{7} c_{5} x^{2 n-25} 2^{5}+^{7}c{ }_{6} x^{n-30}{2 }^{6} \\ +{ }^{7} c_{7} x^{-35} 2^{7}}

Sum of coeffficiet from beginning

\mathrm{s_{1}=1, s_{2}=1+14=15, S_{3}=1+14+84=99 }\\

\mathrm{s_{4}=1+14+84+280=379, S_{5}=1+14+24+280+560=939} \\

\mathrm{So \quad 3 n-20>0 \quad \& \quad 2 n-25 \leqslant 0 }\\

\mathrm{\Rightarrow n>\frac{20}{3} \quad \& \quad n \leqslant \frac{25}{2}} \\

\mathrm{\Rightarrow n=7,8,9,10,11,12} \\

\mathrm{\text { sum }=7+8+9+10+11+12=57}

Hence answer is \mathrm{57}

 

Posted by

Divya Prakash Singh

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