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7.Binomial Theorem
medium
${n^n}{\left( {\frac{{n + 1}}{2}} \right)^{2n}}$ = . . .
A
${\left( {\frac{{n + 1}}{2}} \right)^3}$ કરતાં ઓછું
B
${\left( {\frac{{n + 1}}{2}} \right)^3}$ કરતાં મોટું
C
${(n!)^3}$ કરતાં મોટું
D
$(b)$ અને $(c)$ બંને
Solution
(b) $y = {n^n}{\left( {\frac{{n + 1}}{2}} \right)^{2n}}$
Put $n = 2$, $y = {2^2}{\left( {\frac{3}{2}} \right)^4} = 4\,.\,\frac{{81}}{{8 \times 2}} = \frac{{81}}{4}\tilde – 20$
Option $(a) = {\left( {\frac{{n + 1}}{2}} \right)^3} = \frac{{27}}{8} < y$
Option $(b) = {\left( {\frac{{n + 1}}{2}} \right)^3} = \frac{{27}}{8} < y$
Option $(c) = {(2!)^3} = 8 < y$
Standard 11
Mathematics
