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31/01/2022 1:26 pm
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The sum of 10 terms of the series
\(\frac{3}{1^2 \times 2^2}\) + \(\frac{5}{2^2 \times 3^2}\) + \(\frac{7}{3^2 \times 4^2}\)+....is
(1) 1
(2) \(\frac{120}{121}\)
(3) \(\frac{99}{100}\)
(4) \(\frac{143}{144}\)
1 Answer
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31/01/2022 1:31 pm
Correct answer: (2) \(\frac{120}{121}\)
Explanation:
S = \(\frac{2^2 - 1^2}{1^2 \times 2^2}\) + \(\frac{3^2 - 2^2}{2^2 \times 3^2}\) + \(\frac{4^2 - 3^2}{3^2 \times 4^2}\) + ....
= \(\Big[\frac{1}{1^2} - \frac{1}{2^2}\Big]\) + \(\Big[\frac{1}{2^2} - \frac{1}{3^2}\Big]\) + \(\Big[\frac{1}{3^2} - \frac{1}{4^2}\Big]\) + ....+ \(\Big[\frac{1}{10^2} - \frac{1}{11^2}\Big]\)
= 1 - \(\frac{1}{121}\)
= \(\frac{120}{121}\)