Question

The sum of the first 20 terms of the series 1/5*6+1/6*7+1/7*8+......is

Answer 1

$\frac{1}{5\times 6}+\frac{1}{6\times 7}+\frac{1}{7\times 8}+...upto\;20\;terms=\frac{16}{105}$

Step-by-step explanation:

Series

$\frac{1}{5\times 6}+\frac{1}{6\times 7}+\frac{1}{7\times 8}+...upto\;20\;terms$

n=20

$a_1=\frac{1}{5\times 6}=\frac{1}{5}-\frac{1}{6}$

$a_2=\frac{1}{6\times 7}=\frac{1}{6}-\frac{1}{7}$

$a_3=\frac{1}{7\times 8}=\frac{1}{7}-\frac{1}{8}$

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$a_{n-1}=\frac{1}{(n-1)n}=\frac{1}{n-1}-\frac{1}{n}$

$a_n=\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$

Substitute n=20

$a_{19}=\frac{1}{19}-\frac{1}{20}$

$a_{20}=\frac{1}{20}-\frac{1}{21}$

Adding all term

$a_1+a_2+a_3+...+a_{20}=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...-\frac{1}{20}+\frac{1}{20}-\frac{1}{21}$

$a_1+a_2+a_3+...+a_{20}=\frac{1}{5}-\frac{1}{21}$

$a_1+a_2+a_3+...+a_{20}=\frac{21-5}{105}=\frac{16}{105}$

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