Fractions are a pain

Calculus Level 5

Find the value of 1 7 12 + 25 288 91 5184 + + ( 1 ) r 3 r + 4 r r × 1 2 r + 1 - \frac{7}{12} + \frac{25}{288} - \frac{91}{5184} + \dots + (-1)^r \frac{3^r + 4^r}{r \times 12^r} + \dots

(Note: When r = 0 r=0 , ( 1 ) r 3 r + 4 r r × 1 2 r (-1)^r \frac{3^r + 4^r}{r \times12^r} would be but for the sake of the problem, it's defined to be 1 1 here.)


The answer is 0.489.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Josh Banister
Nov 15, 2015

1 + r = 1 ( 1 ) r 3 r + 4 r r × 1 2 r = 1 + r = 1 ( 1 ) r ( 1 r × 4 r + 1 r × 3 r ) = 1 r = 1 ( 1 ) r ( 1 3 ) r ) r r = 1 ( 1 ) r ( 1 4 ) r r = 1 ln ( 1 + 1 3 ) ln ( 1 + 1 4 ) = 1 ln ( 4 3 ) ln ( 5 4 ) = 1 ln ( 5 3 ) 0.489 \begin{aligned} 1 + \sum_{r=1}^{\infty} (-1)^r \frac{3^r + 4^r}{r \times 12^r} &= 1 + \sum_{r=1}^{\infty} (-1)^r (\frac{1}{r \times 4^r} + \frac{1}{r \times 3^r}) \\ &= 1 - \sum_{r=1}^{\infty} -(-1)^r \frac{( \frac{1}{3})^r )}{r} - \sum_{r=1}^{\infty} -(-1)^r \frac{(\frac{1}{4})^r}{r} \\ &= 1 - \ln (1 + \frac{1}{3}) - \ln (1 + \frac{1}{4}) \\ &= 1 - \ln(\frac{4}{3}) - \ln(\frac{5}{4}) \\ &= 1 - \ln (\frac{5}{3}) \\ &\approx 0.489 \end{aligned}

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...