Infinite Series

[sickplaya]

What is the sum of the following infinite series:

5x^2 + 9x^4 + 13x^6 + 17x^8 +.....
(x=0.85634884)

Round the answer to the nearest whole number.




Please show me all the steps; thanks.

 

[tkhunny]

5x^2 + 9x^4 + 13x^6 + 17x^8 +.....

Just add it all up.

5x^2 + 9x^4 + 13x^6 + 17x^8 +..... = S
5x^4 + 9x^6 + 13x^8 + 17x^10 +..... = S*x^2

Subtracting

5x^2 + 4x^4 + 4x^6 + 4x^8 +..... = S*(1-x^2)

Judicious Factoring

5x^2 + (4x^4)*(1 + x^2 + x^4 +.....) = S*(1-x^2)

Can you add up the one in the parentheses and solve for S?

It's not pretty, but it works.

 

[Denis]

Hint: answer is a 2digit number with both digits the same :shock:

 

[galactus]

This may be little belated but here's a method:

You have \(\displaystyle \sum_{n=1}^{\infty}\(4n+1)x^{2n}\), x<1

=$\displaystyle 4\sum_{n=1}^{\infty}nx^{2n}+\sum_{n=1}^{\infty}x^{2n}$

It can be shown that:

$\displaystyle \sum_{n=1}^{\infty}nx^{2n}=\sum_{n=0}^{\infty}x^{2n}\sum_{n=1}^{\infty}x^{2n}$

$\displaystyle \sum_{n=0}^{\infty}x^{2n}=\frac{1}{1-x^{2}}=\frac{1}{1-(.85634884)^{2}}=3.75=\frac{15}{4}$

Therefore, $\displaystyle \sum_{n=1}^{\infty}x^{2n}=2.75=\frac{11}{4}$

$\displaystyle (4)(3.75)(2.75)=(4)(\frac{15}{4})(\frac{11}{4})=41.25$

So we have: $\displaystyle 41.25+2.75=44$