If n is a natural number then-n-1-2-n- n - is true-when-

Check through option, the condition (n+1/2)^n ≥ n ! is true for n ≥ 1. ...

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If n is a nat...

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If n is a natural number then ${(\frac{n + 1}{2})}^{n}$ ≥ n ! is true

when

n > 1

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$n\geq$ 1

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n > 2

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$n \geq 2$

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{(\frac{n + 1}{2})}^{n} \geq n!

{(\frac{n + 1}{2})}^{n} \geq n!

If n is a natural number then ${(\frac{n + 1}{2})}^{n}$ ≥ n ! is true

when

If n is a natural number then ${(\frac{n + 1}{2})}^{n}$ ≥ n ! is true

when

(1 + x)^{n} = n \sum r = 0^{n} C_{r} x^{n},

\frac{C_{0}}{1 \cdot 2} 2^{2} + \frac{C_{1}}{2 \cdot 3} 2^{3} + \frac{C_{2}}{3 \cdot 4} 2^{4} + \dots + \frac{C_{n}}{(n + 1) (n + 2)} 2^{n + 2}

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