If-5-2-3-5-7-3-3-2-5-7-9-4-3-3-then-find-the-value-of-2-4-

Question Number 22547 by Tinkutara last updated on 20/Oct/17

If α = \frac{5}{2! 3} + \frac{5.7}{3! 3^{2}} + \frac{5.7.9}{4! 3^{3}}, \dots then find

the value of α^{2} + 4 α .

Answered by ajfour last updated on 20/Oct/17

{(1 + x)}^{n} = 1 + n x + \frac{n (n - 1)}{2!} x^{2} + \dots

l e t x = - \frac{2}{3}, n \to - n

{(1 + x)}^{- n} ∣_{x = - \frac{2}{3}} = 1 + \frac{2 n}{3} + \frac{2 n (2 n + 2)}{2! 3^{2}}

+ \frac{2 n (2 n + 2) (2 n + 4)}{3! 3^{3}} + . .

l e t n = 3 / 2

\Rightarrow {(1 - \frac{2}{3})}^{- 3 / 2} = 1 + 1 + \frac{5}{2! 3} + \frac{5.7}{3! 3^{2}} + . .

\Rightarrow 3 \sqrt{3} = 2 + α

α = 3 \sqrt{3} - 2

α^{2} + 4 α = α (α + 4)

= (3 \sqrt{3} - 2) (3 \sqrt{3} + 2)

= 23 .

Commented by ajfour last updated on 20/Oct/17

s e e k i n g α .

Commented by Tinkutara last updated on 20/Oct/17

How you have written x = - \frac{2}{3}, n = \frac{3}{2} ?

Commented by Tinkutara last updated on 03/Jan/18

Thanks Sir!

Commented by ajfour last updated on 03/Jan/18

α n e e d t o b e g e n e r a t e d, 3, 3^{2}, 3^{3}, . .

i n d e n o m i n a t o r c o a x e d m e t o

p u t x = \frac{2}{3}, \dots

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