Question

Solve :—

$\frac{1}{1} + \frac{1}{ 1 + 2} + \frac{1}{1 + 2 + 3} + .. \: .. \: .. + \frac{1}{1 + 2 + 3 + .. \: .. \: .. + 15}$
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Answer 1

Step-by-step explanation:

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Answer 2

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Solve :—

$\frac{1}{1} + \frac{1}{ 1 + 2} + \frac{1}{1 + 2 + 3} + .. \: .. \: .. + \frac{1}{1 + 2 + 3 + .. \: .. \: .. + 15}$

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$➪ T1 = \frac{1}{1} \\ \\ ➪ T2 = \frac{1}{1 + 2} \\ \\ ➪ T3 = \frac{1}{1 + 2 + 3} \\ ⇓ \\ ⇓ \\ ⇓ \\ ➪Tn = \frac{1}{1 + 2 + 3 + .. \: .. \: .. + n} = \\ \\ = \frac{1}{summation \: of \: n} \\ \\ = \frac{2}{n(n + 1)}$

$⇒2[ \frac{1}{n} - \frac{1}{n + 1}]$

⇒S15 = T1 + T2 +T3 +.. .. .. .. .. +T15

$⇒2 [\frac{1}{1} - \frac{1}{2} )2( \frac{1}{2} - \frac{1}{3} ) + 2( \frac{1}{3} - \frac{1}{4} ).. \: .. \: + 2( \frac{1}{15} - \frac{1}{16} ]$

$⇒2[\frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} .. \: .. \: .. + \frac{1}{15} - \frac{1}{16}]$

$⇒2[1 - \frac{1}{16} ]$

$⇒ \cancel2 \times \frac{15}{ \cancel{16} }$

$⇒ \frac{15}{8}$