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in Linear Inequations by (23.5k points)
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If \(\big(\frac{n+1}{2}\big)^n\big(\frac{2n+1}{3}\big)^n >(n!)^k\), then k = 

(a) 1 

(b) 3 

(c) 2 

(d) 4

1 Answer

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Best answer

(c) 2

\(\frac{1^2+2^2+3^2+......n^2}{n}≥(1^2.2^2.3^2......n^2)^{\frac1n}\)

⇒ \(\frac{n(n+1)(2n+1)}{6n}\) > (1.2.3......n)\(\frac2n\)

⇒ \(\big(\frac{n+1}{2}\big)\big(\frac{2n+1}{3}\big) ≥(n!)^{\frac2n}\)

\(\big(\frac{n+1}{2}\big)^n\big(\frac{2n+1}{3}\big)^n ≥(n!)^2\)

k = 2.

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