Find the 13th term in the expansion of 9x- 1 3- x 18-

In a binomial expansion (a+b)^n the r+1th term is ^n C _ r a ^ n r b ^ r here n=18,r=12,a=9x,b = 13x^1/2 therefore by substituing we ...

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Standard XII

Mathematics

Middle Terms

Find the 13...

Question

Find the $13^{t h}$ term in the expansion of ${(9 x - \frac{1}{3 \sqrt{x}})}^{18}$ .

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Solution

in a binomial expansion $(a + b)^{n}$ the $r + 1 t h$ term is $^{n} C_{r} a^{n - r} b^{r}$
here $n = 18, r = 12, a = 9 x, b$ = $\frac{1}{3 x^{1 / 2}}$

therefore by substituing we get $^{18} C_{12} {(9 x)}^{6} {(\frac{- 1}{3 \sqrt{x}})}^{12} =^{18} C_{12} = 18564$

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Find the 13th term in the expansion of (9x−13√x)18.

in a binomial expansion (a+b)n the r+1th term is nCran−rbrhere n=18,r=12,a=9x,b=13x1/2therefore by substituing we get 18C12(9x)6(−13√x)12=18C12=18564

Find the $13^{t h}$ term in the expansion of ${(9 x - \frac{1}{3 \sqrt{x}})}^{18}$ .

in a binomial expansion $(a + b)^{n}$ the $r + 1 t h$ term is $^{n} C_{r} a^{n - r} b^{r}$
here $n = 18, r = 12, a = 9 x, b$ = $\frac{1}{3 x^{1 / 2}}$

therefore by substituing we get $^{18} C_{12} {(9 x)}^{6} {(\frac{- 1}{3 \sqrt{x}})}^{12} =^{18} C_{12} = 18564$