Show that the middle term in the expansion of 1-x2n is 1-3-5-2n-1n- 2nxn- where n is a positive integer-

(1+x)^2n = ^2nC_0x^0+^2nC_1x^1+—+^2nC_2nx^2n = ∑_r = 0^2n x^r. ^2nC_r So, there are (2n+1) terms in the expansion #160;the middle term be for r=nwhi ...

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Standard XII

Mathematics

Middle Terms

Show that the...

Question

Show that the middle term in the expansion of $(1 + x)^{2 n}$ is $\frac{1.3.5... (2 n - 1)}{n!}$ $2^{n} x^{n}$ , where n is a positive integer.

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Solution

$(1 + x)^{2 n} =^{2 n} C_{0} x^{0} +^{2 n} C_{1} x^{1} + - - - +^{2 n} C_{2 n} x^{2 n}$

$= 2 n \sum r = 0 x^{r} .^{2 n} C_{r}$
So, there are (2n+1) terms in the expansion

the middle term be for r=n
which is $^{2 n} C_{n} . x^{n}$
$= \frac{(2 n)!}{n! (2 n - n)!} x^{n}$
$= \frac{2 n . (2 n - 1) (2 n - 2) - - - 4.3.2.1}{n! . n!} x^{n}$

$= \frac{(1.3.5 - - - (2 n - 1)) 2^{n} (1.2.3... n)}{n! n!}, x^{n} .$

$= \frac{1.3.5 - - - (2 n - 1)}{n!} 2^{n} x^{n} .$

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Show that the middle term in the expansion of (1+x)2n is 1.3.5...(2n−1)n! 2nxn, where n is a positive integer.

(1+x)2n=2nC0x0+2nC1x1+−−−+2nC2nx2n=2n∑r=0xr.2nCrSo, there are (2n+1) terms in the expansion the middle term be for r=nwhich is 2nCn.xn=(2n)!n!(2n−n)!xn=2n.(2n−1)(2n−2)−−−4.3.2.1n!.n!xn=(1.3.5−−−(2n−1))2n(1.2.3...n)n!n!,xn.=1.3.5−−−(2n−1)n!2nxn.

Show that the middle term in the expansion of $(1 + x)^{2 n}$ is $\frac{1.3.5... (2 n - 1)}{n!}$ $2^{n} x^{n}$ , where n is a positive integer.