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7.Binomial Theorem
medium
${n^n}{\left( {\frac{{n + 1}}{2}} \right)^{2n}}$ होगा
A
$< {\left( {\frac{{n + 1}}{2}} \right)^3}$
B
$> {\left( {\frac{{n + 1}}{2}} \right)^3}$
C
$>{(n!)^3}$
D
$(b)$ and $(c)$
Solution
$y = {n^n}{\left( {\frac{{n + 1}}{2}} \right)^{2n}}$
$n = 2$ रखने पर, $y = {2^2}{\left( {\frac{3}{2}} \right)^4} = 4\,.\,\frac{{81}}{{8 \times 2}} = \frac{{81}}{4} \tilde – 20]$
विकल्प $(a)$ से $ = {\left( {\frac{{n + 1}}{2}} \right)^3} = \frac{{27}}{8} < y$
विकल्प $(b)$ से $ = {\left( {\frac{{n + 1}}{2}} \right)^3} = \frac{{27}}{8} < y$
विकल्प $(c)$ से $ = {(2!)^3} = 8 < y$
Std 11
Mathematics
