If-n-is-even-positive-integer-then-the-condition-that-the-greatest-term-in-the-expansion-of-1-x-n-may-have-the-greatest-coefficient-also-is-

Question Number 63860 by gunawan last updated on 10/Jul/19

If n is even positive integer, then the

condition that the greatest term in the

expansion of {(1 + x)}^{n} may have the

greatest coefficient also is

Answered by mr W last updated on 10/Jul/19

T_{r} = C_{r}^{n} x^{r}

n = 2 m

T_{m - 1} = C_{m - 1}^{2 m} x^{m - 1}

T_{m} = C_{m}^{2 m} x^{m} s h o u l d b e m a x i m u m .

T_{m + 1} = C_{m + 1}^{2 m} x^{m + 1}

\frac{T_{m}}{T_{m - 1}} = \frac{C_{m}^{2 m}}{C_{m - 1}^{2 m}} x > 1

\frac{(2 m)! (2 m - m + 1)! (m - 1)!}{(2 m - m)! m! (2 m)!} x > 1

\frac{(m + 1)}{m} x > 1

\frac{(n + 2) x}{n} > 1

\Rightarrow x > \frac{n}{n + 2} = 1 - \frac{2}{n + 2}

\frac{T_{m}}{T_{m + 1}} = \frac{C_{m}^{2 m}}{{x C}_{m + 1}^{2 m}} > 1

\frac{n + 2}{n x} > 1

\frac{1}{x} > \frac{n}{n + 2}

\Rightarrow x < \frac{n + 2}{n} = 1 + \frac{2}{n}

\Rightarrow c o n d i t i o n i s

1 - \frac{2}{n + 2} < x < 1 + \frac{2}{n}

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