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  4. Sum of the series ( 100 C 1)2+2( 100 C 2)2+3( 100 C 3)2+ ldots ldots . .+100( 100 C 100)2 equals

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Q. Sum of the series $\left({ }^{100} C _1\right)^2+2\left({ }^{100} C _2\right)^2+3\left({ }^{100} C _3\right)^2+\ldots \ldots . .+100\left({ }^{100} C _{100}\right)^2$ equals

Binomial Theorem

Solution:

$ S =0 \cdot\left({ }^{100} C _0\right)^2+1 \cdot\left({ }^{100} C _1\right)^2+2 \cdot\left({ }^{100} C _2\right)^2+3 \cdot\left({ }^{100} C _3\right)^2+\ldots \ldots \ldots+100 \cdot\left({ }^{100} C _{100}\right)^2 $
$S =100 \cdot\left({ }^{100} C _0\right)^2+99 \cdot\left({ }^{100} C _1\right)^2+\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots+0 . \ldots \ldots \ldots+\left({ }^{100} C _{100}\right)^2 $
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$2 S =100\left[\left({ }^{100} C _0\right)^2+\left({ }^{100} C _1\right)^2+\left({ }^{100} C _2\right)^2+\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots+\left({ }^{100} C _{100}\right)^2\right] $
$2 S =100 \cdot{ }^{200} C _{100}=100 \cdot \frac{(200) !}{100 ! \cdot 100 !} \Rightarrow 2 S =\frac{100 \cdot 200 \cdot(199) !}{100 \cdot(99) ! \cdot(100) !} \Rightarrow S =100 \cdot{ }^{199} C _{99}$