dear student LHS=(2n-2)!n!(n-2)!+2(2n-2)!(n-1)!(n-1)!+(2n-2)!n!(n-2)!=2(2n-2)!(n-1)!(n-2)!1n+1n-1=2(2n-2)!(n-1)!(n-2)!(2n-1)n(n-1)=2(2n-1)!(n-1)!(n)!=2 n2n-1Cso using mathematical induction, we prove2 n2n-1C>4nn+1 for n>2n2n-1C>2nn+1 for n>2for n=336-1C>63+110>3/2assuming true for n=n alsofor n=n+1LHS=n+12(n+1)-1C=n+12n+1C=(2n+1)!n!(n+1)!=(2n+1)(2n)(2n-1)!n!(n+1)n(n-1)!=(2n+1)(2n)(n+1)nn2n-1C=(4n+2)(n+1)n2n-1Cclearly 4n+2n+1>4nn+1so he inequality is also true for n+1hence proved regards

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dear student

L H S = \frac{(2 n - 2)!}{n! (n - 2)!} + \frac{2 (2 n - 2)!}{(n - 1)! (n - 1)!} + \frac{(2 n - 2)!}{n! (n - 2)!} = \frac{2 (2 n - 2)!}{(n - 1)! (n - 2)!} (\frac{1}{n} + \frac{1}{n - 1}) = \frac{2 (2 n - 2)!}{(n - 1)! (n - 2)!} \frac{(2 n - 1)}{n (n - 1)} = \frac{2 (2 n - 1)!}{(n - 1)! (n)!} = 2 ._{n}^{2 n - 1} C s o u \sin g m a t h e m a t i c a l i n d u c t i o n, w e p r o v e 2 ._{n}^{2 n - 1} C > \frac{4 n}{n + 1} f o r n > 2_{n}^{2 n - 1} C > \frac{2 n}{n + 1} f o r n > 2 f o r n = 3_{3}^{6 - 1} C > \frac{6}{3 + 1} 10 > 3 / 2 a s s u m i n g t r u e f o r n = n a l s o f o r n = n + 1 L H S =_{n + 1}^{2 (n + 1) - 1} C =_{n + 1}^{2 n + 1} C = \frac{(2 n + 1)!}{n! (n + 1)!} = \frac{(2 n + 1) (2 n) (2 n - 1)!}{n! (n + 1) n (n - 1)!} = {\frac{(2 n + 1) (2 n)}{(n + 1) n}}_{n}^{2 n - 1} C = {\frac{(4 n + 2)}{(n + 1)}}_{n}^{2 n - 1} C c l e a r l y \frac{4 n + 2}{n + 1} > \frac{4 n}{n + 1} s o h e i n e q u a l i t y i s a l s o t r u e f o r n + 1 h e n c e p r o v e d

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