If the sum of the coefficients of all the positive powers of x, in the Binomial expansion of ${\left( {{x^n} + {2 \over {{x^5}}}} \right)^7}$ is 939, then the sum of all the possible integral values of n is _________.
Answer (integer)
57
Solution
<p>Given, Binomial expression is</p>
<p>$= {\left( {{x^n} + {2 \over {{x^5}}}} \right)^7}$</p>
<p>$\therefore$ General term</p>
<p>$${T_{r + 1}} = {}^7{C_r}\,.\,{({x^n})^{7 - r}}\,.\,{\left( {{2 \over {{x^5}}}} \right)^r}$$</p>
<p>$= {}^7{C_r}\,.\,{x^{7n - nr - 5r}}\,.\,{2^r}$</p>
<p>For positive power of x,</p>
<p>$7n - nr - 5r > 0$</p>
<p>$\Rightarrow 7n > r(n + 5)$</p>
<p>$\Rightarrow r < {{7n} \over {n + 5}}$</p>
<p>As r represent term of binomial expression so r is always integer.</p>
<p>Given that sum of coefficient is 939.</p>
<p>When $r = 0$,</p>
<p>sum of coefficient $= {}^7{C_0}\,.\,{2^0} = 1$</p>
<p>when $r = 1$,</p>
<p>sum of coefficient $= {}^7{C_0}\,.\,{2^0} + {}^7{C_1}\,.\,{2^1} = 1 + 14 = 15$</p>
<p>when $r = 2$,</p>
<p>sum of coefficient</p>
<p>$= {}^7{C_0}\,.\,{2^0} + {}^7{C_1}\,.\,{2^1} + {}^7{C_2}\,.\,{2^2}$</p>
<p>$= 1 + 14 + 84$</p>
<p>$= 99$</p>
<p>when $r = 3$,</p>
<p>sum of coefficient</p>
<p>$$ = {}^7{C_0}\,.\,{2^0} + {}^7{C_1}\,.\,{2^1} + {}^7{C_2}\,.\,{2^2} + {}^7{C_3}\,.\,{2^3}$$</p>
<p>$= 1 + 14 + 84 + 280$</p>
<p>$= 379$</p>
<p>when $r = 4$,</p>
<p>sum of coefficient</p>
<p>$$ = {}^7{C_0}\,.\,{2^0} + {}^7{C_1}\,.\,{2^1} + {}^7{C_2}\,.\,{2^2} + {}^7{C_3}\,.\,{2^3} + {}^7{C_4}\,.\,{2^4}$$</p>
<p>$= 1 + 14 + 84 + 280 + 560$</p>
<p>$= 939$</p>
<p>$\therefore$ For r = 4 sum of coefficient = 939</p>
<p>To get value of r = 4, value of ${{7n} \over {n + 5}}$ should be between 4 and 5.</p>
<p>$\therefore$ $4 < {{7n} \over {n + 5}} < 5$</p>
<p>$\Rightarrow 4n + 20 < 7n < 5n + 25$</p>
<p>$\therefore$ $4n + 20 < 7n$</p>
<p>$\Rightarrow 3n > 20$</p>
<p>$\Rightarrow n > {{20} \over 3}$</p>
<p>$\Rightarrow n > 6.66$</p>
<p>and</p>
<p>$7n < 5n + 25$</p>
<p>$\Rightarrow 2n < 25$</p>
<p>$\Rightarrow n < 12.5$</p>
<p>$\therefore$ $6.66 < n < 12.5$</p>
<p>$\therefore$ Possible integer values of $n = 7,8,9,10,11,12$</p>
<p>$\therefore$ Sum of values of $n = 7 + 8 + 9 + 10 + 11 + 12$</p>
<p>$= 57$</p>
About this question
Subject: Mathematics · Chapter: Binomial Theorem · Topic: Binomial Expansion
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