Sequences and series › Infinite series

Alternating fraction sums

Solve the first problem 18 problems

These are 9 of the 18 problems. Sign in for free to see them all. Sign in

  1. Sum of squares divided by 365 ●○○○○ 10 ptsWhat is the value of the expression 102+112+122+132+142365\frac{10^{2}+11^{2}+12^{2}+13^{2}+14^{2}}{365}?
  2. Ratio of alternating sums ●○○○○ 10 ptsFind the value of 2−4+6−8+10−12+143−6+9−12+15−18+21\frac{2-4+6-8+10-12+14}{3-6+9-12+15-18+21} as a fraction in lowest terms. Give the numerator and the denominator.
  3. Ratio of alternating power-of-two sums ●○○○○ 10 ptsCompute 2+4−8+16+32−644+8−16+32+64−128\dfrac{2+4-8+16+32-64}{4+8-16+32+64-128}. Give your answer as a decimal.
  4. Sum of fractions with shared denominators ●●○○○ 20 ptsFind the value of (12+13+⋯+120)+(23+⋯+220)+⋯+(1819+1820)+1920\left(\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{20}\right)+\left(\frac{2}{3}+\cdots+\frac{2}{20}\right)+\cdots+\left(\frac{18}{19}+\frac{18}{20}\right)+\frac{19}{20}.
  5. Telescoping sum of odd-numerator fractions ●●○○○ 20 ptsCalculate 34+536+7144+9400+11900+131764+153136\frac{3}{4}+\frac{5}{36}+\frac{7}{144}+\frac{9}{400}+\frac{11}{900}+\frac{13}{1764}+\frac{15}{3136}. Give your answer as a decimal.
  6. Nested fraction sum with halving ●●○○○ 20 ptsDetermine the value of 2002+12(2001+12(2000+⋯+12(3+12⋅2))⋯ )2002+\frac{1}{2}\left(2001+\frac{1}{2}\left(2000+\cdots+\frac{1}{2}\left(3+\frac{1}{2}\cdot 2\right)\right)\cdots\right).
  7. Ratio of weighted harmonic sums ●●●○○ 50 ptsDetermine the value of 20161+20152+20143+⋯+1201612+13+14+⋯+12017\frac{\frac{2016}{1} + \frac{2015}{2} + \frac{2014}{3} + \dots + \frac{1}{2016}}{\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots + \frac{1}{2017}}.
  8. Nested half-fraction sum ●●●○○ 50 ptsFind the value of 2002+12(2001+12(2000+⋯+12(3+12⋅2))⋯ )2002 + \frac{1}{2} \left( 2001 + \frac{1}{2} \left( 2000 + \dots + \frac{1}{2} \left( 3 + \frac{1}{2} \cdot 2 \right) \right) \dotsb \right).
  9. Alternating-parity sum of reciprocal squares ●●●○○ 50 ptsCompute the value of 1+132+152+172+⋯+19972+19992−110022−110042−110062−⋯−119982−1200021+122+132+142+⋯+19992+110002\frac{1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\cdots+\frac{1}{997^2}+\frac{1}{999^2}-\frac{1}{1002^2}-\frac{1}{1004^2}-\frac{1}{1006^2}-\cdots-\frac{1}{1998^2}-\frac{1}{2000^2}}{1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots+\frac{1}{999^2}+\frac{1}{1000^2}} without using a calculator. Give your answer as a decimal.

More in Infinite series