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Question Number 4639 by FilupSmith last updated on 17/Feb/16

Is the section marked (*) c o r r e c t ?

\underset{i = 1}{\prod^{x}} i = x!

\underset{n = 1}{\prod^{x}} (\underset{i = 1}{\prod^{n}} i) = 1! \times 2! \times \dots \times x!

= 1 \times (1 \times 2) \times (1 \times 2 \times 3) \times \dots \times (1 \times 2 \times \dots \times x)

= 1^{x} 2^{x - 1} 3^{x - 2} \dots {(x - 1)}^{2} x^{1}

(*)

\underset{m = 1}{\prod^{x}} (\underset{n = 1}{\prod^{m}} (\underset{i = 1}{\prod^{n}} i)) = \underset{m = 1}{\prod^{x}} (1^{m} 2^{m - 1} \dots {(m - 1)}^{2} m)

= 1 \times (1^{2} 2^{1}) \times (1^{3} 2^{2} 3^{1}) \times (1^{4} 2^{3} 3^{2} 4^{1}) \times \dots \times (1^{x} 2^{x - 1} \dots x^{1})

= 1^{\underset{i = x - 0}{\sum^{x}} i} \times 2^{\underset{i = x - 1}{\sum^{x}} i} \times 3^{\underset{i = x - 2}{\sum^{x}} i} \times \dots {(x - 1)}^{\underset{i = x - (x - 1) + 1}{\sum^{x}} i} \times x^{\underset{i = x - x + 1}{\sum^{x}} i}

= \underset{i = 1}{\prod^{x}} (i^{(\underset{k = x - i + 1}{\sum^{x}} k)})

= \underset{i = 1}{\prod^{x}} i^{\frac{1}{2} x (2 x - i + 1)}

= \underset{i = 1}{\prod^{x}} i^{x^{2} + \frac{1}{2} (1 - i)}

= \underset{i = 1}{\prod^{x}} i^{(x^{2})} \sqrt{i^{1 - i}}

∴ \underset{m = 1}{\prod^{x}} (\underset{n = 1}{\prod^{m}} (\underset{i = 1}{\prod^{n}} i)) = \underset{i = 1}{\prod^{x}} i^{\frac{1}{2} x (2 x - i + 1)}

Commented by Yozzii last updated on 17/Feb/16

H o w i s 1 \times 1^{2} \times 1^{3} \times \dots \times 1^{x - 1} \times 1^{x} = 1^{\underset{i = x - 0}{\sum^{x}} i} ?

I^{'} m o n l y q u e s t i o n i n g t h e p a t t e r n

a l t h o u g h t h e e q u a t i o n i s r i g h t .

Commented by FilupSmith last updated on 18/Feb/16

Because :

a^{x} \times a^{y} = a^{x + y}

∴ 1 \times 1^{2} \times 1^{3} \times \dots \times 1^{x} = 1^{1 + 2 + 3 + \dots + x}

= 1^{\underset{i = 1}{\sum^{x}} i}

Commented by FilupSmith last updated on 18/Feb/16

1 \times (1^{2} 2^{1}) \times (1^{3} 2^{2} 3^{1}) \times (1^{4} 2^{3} 3^{2} 4^{1}) \times \dots \times (1^{x} 2^{x - 1} \dots x^{1})

= (1 \times 1^{2} \times \dots 1^{x}) \times (2^{2} \times \dots \times 2^{x}) \times (3^{3} \times \dots \times 3^{x}) \dots

f o r 1^{\underset{k = 1}{\sum^{x}} k} = 1^{1} = 1^{\underset{k = 1}{\sum^{x}} k} = 1^{\underset{k = 1}{\sum^{x}} k}

f o r 2^{\underset{k = 2}{\sum^{x}} k} = 2^{\underset{k = 2}{\sum^{x}} k} = 2^{2 + 3 + \dots + x}

But \dots

\underset{m = 1}{\prod^{x}} (\underset{n = 1}{\prod^{m}} (\underset{i = 1}{\prod^{n}} i)) The red part changes the

value for x for each section :

\underset{x = 1}{1} \times \underset{x = 2}{(1^{2} 2^{1})} \times \underset{x = 3}{(1^{3} 2^{2} 3^{3})} \times \dots \times \underset{x = x}{(1^{x} 2^{x - 1} \dots x)}

Commented by FilupSmith last updated on 18/Feb/16

i was tired when i wrote this post

so i dont remember my logic

Commented by FilupSmith last updated on 18/Feb/16

After re reading this i have concluded

that the answer is incorrect and should be :

= \underset{i = 1}{\prod^{x}} i^{\underset{k = 1}{\sum^{x - i + 1}} k}

= \underset{i = 1}{\prod^{x}} i^{\frac{1}{2} (x - i + 1) (x - i + 2)}