please-calculate-A-and-B-A-1-1-4-1-1-9-1-1-16-1-1-4-084-441-and-2021-2-4084441-B-1-2-2-2-3-2-4-2-5-2-6-2-7-2-8-2-9-2-10-2-11-2-

Question Number 161377 by stelor last updated on 17/Dec/21

p l e a s e c a l c u l a t e A a n d B .

A = (1 - \frac{1}{4}) (1 - \frac{1}{9}) (1 - \frac{1}{16}) \dots (1 - \frac{1}{4 084 441}) (a n d 2021^{2} = 4084441)

B = (1^{2} - 2^{2} - 3^{2} + 4^{2}) + (5^{2} - 6^{2} - 7^{2} + 8^{2}) + (9^{2} - 10^{2} - 11^{2} + 12^{2}) + \dots + (2021^{2} - 2022^{2} - 2023^{2} + 2024^{2})

Commented by MJS_new last updated on 17/Dec/21

please check B

(1^{2} - 2^{2} - 3^{2} - 4^{2}) + (4^{2} - 5^{2} - 6^{2} + 7^{2}) + (8^{2} \dots

Commented by stelor last updated on 17/Dec/21

s o r r i e i h a v e m i s t a k e n .

Answered by MJS_new last updated on 17/Dec/21

A_{n} = \underset{j = 2}{\prod^{n}} \frac{j^{2} - 1}{j^{2}} = \frac{n + 1}{2 n} \Rightarrow A_{2021} = \frac{1011}{2021}

Answered by MohammadAzad last updated on 17/Dec/21

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

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

= \underset{n = 2}{\prod^{x}} \frac{n - 1}{n} . \underset{n = 2}{\prod^{x}} \frac{n + 1}{n} = \frac{1}{x} . \frac{x + 1}{2}

= \frac{x + 1}{2 x} (a m a z i n g e h ?)

Y o u k n o w {w h a t}^{'} s m o r e a m a z i n g

\lim_{x \to \infty} \underset{n = 2}{\prod^{x}} (1 - \frac{1}{n^{2}}) = \frac{1}{2}

Answered by MJS_new last updated on 17/Dec/21

B

n^{2} - {(n + 1)}^{2} - {(n + 2)}^{2} + {(n + 3)}^{2} = 4

Commented by stelor last updated on 17/Dec/21

o k a y . t h a n k s

4 \times 2021

Commented by MohammadAzad last updated on 17/Dec/21

C h e c k y o u r s o l u t i o n i t i s

a c t u a l l y 4 \times \frac{2024}{4} = 2024

\underset{i = 0}{\sum^{\frac{n - 1}{4}}} [{(4 i + 1)}^{2} - {(4 i + 1 + 1)}^{2} - {(4 i + 1 + 2)}^{2} + {(4 i + 1 + 3)}^{2}] = n + 3

Commented by MJS_new last updated on 17/Dec/21

{you}^{'} re right

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