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Magical Mathematics[Interesting Approach]> find sum 1^2C1+3^2C3+5^2*C5.......

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find sum 1^2C1+3^2C3+5^2*C5....

Leo Pixie , 11 Years ago

Grade 9

1 Answers

Arun Kumar

Hello Student,

$\\$by simple substitution in $(1+x)^n$we can find$ \\\sum_{r=1}^{n}r^2._{r}^{n}\textrm{C}=n(n-1)2^{n-2}+n2^{n-1} \\\sum_{r=1}^{n}(-1)^{r-1}r^2._{r}^{n}\textrm{C}=0 \\$on adding we get$ \\2(1*_{1}^{n}\textrm{C}+3^2._{3}^{n}\textrm{C}+5^2._{5}^{n}\textrm{C}....)=n(n+1)2^{n-2}$

Thanks & Regards
Arun Kumar
Btech, IIT Delhi
Askiitians Faculty

Last Activity: 11 Years ago

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