Find-1-1-1-2-1-1-2-3-1-1-2-3-40-

Question Number 207845 by hardmath last updated on 28/May/24

Find :

1 + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + \dots + \frac{1}{1 + 2 + 3 + \dots + 40}

Answered by Frix last updated on 28/May/24

\underset{j = 1}{\sum^{n}} (\underset{k = 1}{\sum^{j}} k) = 2 - \frac{2}{n + 1}

n = 40 \Rightarrow answer is \frac{80}{41}

Commented by hardmath last updated on 28/May/24

thank you professor

Answered by MM42 last updated on 28/May/24

= \frac{2}{1 \times 2} + \frac{2}{2 \times 3} + \frac{2}{3 \times 4} + \dots + \frac{2}{39 \times 40} + \frac{2}{40 \times 41}

= 2 [(\frac{1}{1} - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \dots + (\frac{1}{39} - \frac{1}{40}) + (\frac{1}{40} - \frac{1}{41})]

= 2 (1 - \frac{1}{41}) = \frac{80}{41} ✓

Commented by hardmath last updated on 28/May/24

thank you dear professor