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7.Binomial Theorem
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यदि ${a_k} = \frac{1}{{k(k + 1)}},$ जबकि $k = 1,\,2,\,3,\,4,.....,\,n$, तब ${\left( {\sum\limits_{k = 1}^n {{a_k}} } \right)^2} = $
A
$\left( {\frac{n}{{n + 1}}} \right)$
B
${\left( {\frac{n}{{n + 1}}} \right)^2}$
C
${\left( {\frac{n}{{n + 1}}} \right)^4}$
D
${\left( {\frac{n}{{n + 1}}} \right)^6}$
Solution
$\sum\limits_{k = 1}^n {{a_k}} = \sum\limits_{k = 1}^n {\frac{1}{{k\,(k + 1)}}} $
=$\left( {1 – \frac{1}{2}} \right) + \left( {\frac{1}{2} – \frac{1}{3}} \right) + … + \left( {\frac{1}{n} – \frac{1}{{n + 1}}} \right)$
= $1 – \frac{1}{{n + 1}} = \frac{n}{{n + 1}}$
${\left( {\sum\limits_{k = 1}^n {{a_k}} } \right)^2} = {\left( {\frac{n}{{n + 1}}} \right)^2}$.
Standard 11
Mathematics
