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Prove that .^100 C0^(100)C2+^(100)C2^(10...

Prove that `.^100 C_0^(100)C_2+^(100)C_2^(100)C_4+^(100)C_4^(100)C_6++^(100)C_(98)^(100)C_(100)=1/2[.^(200)C_(98)-^(100)C_(49)]dot`

Text Solution

Verified by Experts

To find
`S = .^(100)C_(0).^(100)C_(2)+.^(100)C_(2).^(100)C_(4)+.^(100)C_(4).^(100)C_(6)+"....."+.^(100)C_(98).^(100)C_(100)`
Consider,
`.^(100)C_(0).^(100)C_(2)+.^(100)C_(1).^(100)C_(3) + .^(100)C_(2).^(100)C_(4)+.^(100)C_(3)+.^(100)C_(5)+"...." + .^(100)C_(98).^(100)C_(100)`
`= .^(100)C_(0).^(100)C_(98)+.^(100)C_(1).^(100)C_(97) + .^(100)C_(2).^(100)C_(96)+.^(100)C_(3).^(100)C_(95) + "....."+^(100)C_(98) .^(100)C_(0)`
= Coefficients of `x^(98)` in `(1+x)^(100) (1+x)^(100)`
`= .^(200)C_(98) " "(1)`
Also,
`.^(100)C_(0).^(100)C_(2)-.^(100)C_(1).^(100)C_(3) +.^(100)C_(2).^(100)C_(4)-.^(100)C_(3).^(100)C_(5)+"..."+.^(100)C_(98).^(100)C_(100)`
`=` Cefficient of `x^(98)` in `(1+x)^(100)(1-x)^(100)`
`=` Coefficient of `x^(98)` in `(1-x^(2))^(100)`
`= -.^(100)C_(49)`
Adding (1) and (2), we have
`2(.^(100)C_(0).^(100)C_(2)+.^(100)C_(2).^(100)C_(4)+.^(100)C_(4).^(100)C_(6)+"...."+.^(100)C_(98).^(100)C_(100))`
`= [.^(200)C_(98)-.^(100)C_(149)]`
`rArr .^(100)C_(0).^(100)C_(2)+.^(100)C_(2).^(100)C_(4)+.^(100)C_(4).^(100)C_(6)+"....."+.^(100)C_(98).^(100)C_(100)`
`= 1/2[.^(200)C_(98) - .^(100)C_(49)]`
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