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Question Number 121754 by bemath last updated on 11/Nov/20

	Answered by liberty last updated on 11/Nov/20

	$let y = {(2020!)}^{2}$ $\Rightarrow \ln y = 2 (\ln 2020 + \ln 2019 + \ln 2018 + . . . + \ln 1)$ $let x = 2020^{2020}$ $\Rightarrow \ln x = 2020 \ln (2020)$ $consider \ln x - \ln y = 2020 \ln (2020) - 2 (\ln 2020 + \ln 2019 + . . . + \ln 1)$ $\ln x - \ln y = 2018 . \ln 2020 - 2 \ln (2019!) > 0$ $so \ln x > \ln y \Rightarrow x > y$
	Commented by MJS_new last updated on 11/Nov/20

	${(1!)}^{2} = 1^{} 1^{1} = 1$ ${(2!)}^{2} = 4 2^{2} = 4$ ${(3!)}^{2} = 36 3^{3} = 27$ ${(4!)}^{2} = 576^{} 4^{4} = 256$ $. . .$ ${(n!)}^{2} ⩾ n^{n} \forall n \in N^{★}$ $using {Stirling}^{'} s Approximation$ $\lim_{n \to \infty} (n! - \frac{n^{n}}{e^{n}} \sqrt{2 π n}) = 0 and n! > \frac{n^{n}}{e^{n}} \sqrt{2 π n}$ ${(n!)}^{2} <> n^{n}$ $\frac{n^{2 n}}{e^{2 n}} 2 π n <> n^{n}$ $n^{2 n} 2 π n <> n^{n} e^{2 n}$ $both sides > 0$ $2 n \ln n + \ln n + \ln 2 π <> n \ln n + 2 n$ $(n + 1) \ln n <> 2 n - \ln 2 π$ $\ln n <> 2 - \frac{2 + \ln 2 π}{n + 1}$ $obviously \ln n > 2 \forall n > e^{2}$ $[\ln n = 2 - \frac{2 + \ln 2 π}{n + 1} \Rightarrow n \approx 2.34940]$ $2020 > e^{2} \Rightarrow {(2020!)}^{2} > \frac{2020^{4040}}{e^{4040}} 4040 π > 2020^{2020}$
	Answered by TANMAY PANACEA last updated on 11/Nov/20
	$(n!)^2 =A B=n^n n!=n(n−1)(n−2)(n−3)..(n−r+1)..3×2×1⇚r_(th) term=(n−r+1)=T_r n!=1×2×3×...×r×(n−2)(n−1)n⇚ r_(th) term=r=t_r (n!)^2 =(n×1)×{(n−1)×2}{(n−2)×3...{(n−r+1)r}...{2×(n−1)}{1×n} now look r_(th) term of (n!)^2 is (n−r+1)r value of (T_r t_r −n) is T_r t_r −n (n−r+1)r−n =nr−r^2 +r−n =r(n−r)−1(n−r) =(n−r)(r−1) is greater than zero when n>r>1 so T_r t_r >n T_1 t_1 >n T_2 t_2 >n T_3 t_3 >n ... .... T_n t_n >n multiply both side (T_1 t_1 )(T_2 t_2 )...(T_n t_n )>n^n (n!)^2 >n^n so (2020!)^2 >(2020)^(2020)$
	${(n!)}^{2} = A B = n^{n}$ $n! = n (n - 1) (n - 2) (n - 3) . . (n - r + 1) . . 3 \times 2 \times 1 ⇚ r_{t h} t e r m = (n - r + 1) = T_{r}$ $n! = 1 \times 2 \times 3 \times . . . \times r \times (n - 2) (n - 1) n ⇚ r_{t h} t e r m = r = t_{r}$ ${(n!)}^{2} = (n \times 1) \times {(n - 1) \times 2} {(n - 2) \times 3 . . . {(n - r + 1) r} . . . {2 \times (n - 1)} {1 \times n}$ $n o w l o o k r_{t h} t e r m o f {(n!)}^{2} i s (n - r + 1) r$ $v a l u e o f (T_{r} t_{r} - n) i s$ $T_{r} t_{r} - n$ $(n - r + 1) r - n$ $= n r - r^{2} + r - n$ $= r (n - r) - 1 (n - r)$ $= (n - r) (r - 1) i s g r e a t e r t h a n z e r o w h e n n > r > 1$ $s o T_{r} t_{r} > n$ $T_{1} t_{1} > n$ $T_{2} t_{2} > n$ $T_{3} t_{3} > n$ $. . .$ $. . . .$ $T_{n} t_{n} > n$ $m u l t i p l y b o t h s i d e$ $(T_{1} t_{1}) (T_{2} t_{2}) . . . (T_{n} t_{n}) > n^{n}$ ${(n!)}^{2} > n^{n}$ $s o {(2020!)}^{2} > {(2020)}^{2020}$
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