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I have problem in Binomial Theorem Question:

Find coefficient of $x^0$ term $$[9x-\frac{b}{c\cdot x^2}]^{17}$$

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2 Answers 2

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$0$ because you need to use $2N$ terms of $9x$ and $N$ terms of $1/x^2$ to balance the powers. In total there are $17$ terms which is not devicible by $3$

You can also simply expand :)

$$ -\frac{b^{17}}{c^{17} x^{34}}+\frac{153 b^{16}}{c^{16} x^{31}}-\frac{11016 b^{15}}{c^{15} x^{28}}+\frac{495720 b^{14}}{c^{14} x^{25}}-\frac{15615180 b^{13}}{c^{13} x^{22}}+\frac{365395212 b^{12}}{c^{12} x^{19}}-\frac{6577113816 b^{11}}{c^{11} x^{16}}+\frac{93019181112 b^{10}}{c^{10} x^{13}}-\frac{1046465787510 b^9}{c^9 x^{10}}+\frac{9418192087590 b^8}{c^8 x^7}-\frac{67810983030648 b^7}{c^7 x^4}+\frac{388371993720984 b^6}{c^6 x}-\frac{1747673971744428 b^5 x^2}{c^5}+\frac{6049640671423020 b^4 x^5}{c^4}-\frac{15556218869373480 b^3 x^8}{c^3}+\frac{28001193964872264 b^2 x^{11}}{c^2}-\frac{31501343210481297 b x^{14}}{c}+16677181699666569 x^{17}$$

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  • $\begingroup$ But answer in my book is $$12C_4[\frac{b}{c}]^{4}\cdot9^{8}$$ $\endgroup$
    – Ganesh
    Sep 17, 2015 at 11:14
  • $\begingroup$ which is probably the correct answer for the power $12$ not $17$. What is your notation for $C_4$? For $12$ one should get $\frac{21308126895 b^4}{c^4}$ $\endgroup$
    – Nikolay Gromov
    Sep 17, 2015 at 11:18
  • $\begingroup$ @Nickolay But question in my book is 17 $\endgroup$
    – Ganesh
    Sep 17, 2015 at 11:19
  • $\begingroup$ Write to a publisher, because the answer is given for $12$, what can I do... $\endgroup$
    – Nikolay Gromov
    Sep 17, 2015 at 11:20
  • $\begingroup$ C is combination $\endgroup$
    – Ganesh
    Sep 17, 2015 at 11:22
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Notice, Binomial expansion $$\left(9x-\frac{b}{cx^2}\right)^{17}$$ $$=^{17}C_{0}(9x)^{17}\left(\frac{b }{cx^2}\right)^{0}+^{17}C_{1}(9x)^{16}\left(\frac{b }{cx^2}\right)^{1}+^{17}C_{2}(9x)^{15}\left(\frac{b }{cx^2}\right)^{2}+\ldots + ^{17}C_{17}(9x)^{0}\left(\frac{b }{cx^2}\right)^{17}$$

$r^{th}$ term is given as $$T_r=^{17}C_{r-1}(9x)^{17-r+1}\left(\frac{b }{cx^2}\right)^{r-1}$$$$=^{17}C_{r-1}(9x)^{17-r+1}\left(\frac{b }{cx^2}\right)^{r-1}$$ $$=^{17}C_{r-1}\left(\frac{b }{c}\right)^{r-1}(x)^{18-r-2r+2}$$ $$=^{17}C_{r-1}\left(\frac{b }{c}\right)^{r-1}(x)^{20-3r}$$ For $x^0$ term, the power of $x$ in $T_r$ should be zero hence let's put $$20-3r=0\iff r=\frac{20}{3}$$ but $r$ is an integer hence $x^0$ term (constant) is not there in the expansion.

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  • $\begingroup$ In your calculation, where did the 9 in 9x go? $\endgroup$
    – Björn Lindqvist
    Jan 15, 2017 at 16:48

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