Binomial TheoremVery Hard
Question
If Cr stands for nCr, then the sum of the series
[Co2 - 2C12 + 3C22 - .... + (- 1)r (n + 1)Cn2]
where n is an even positive integer, is equal to
[Co2 - 2C12 + 3C22 - .... + (- 1)r (n + 1)Cn2] where n is an even positive integer, is equal to
Options
A.(-1)n/2(n + 2)
B.(-1)n(n + 1)
C.(-1)n/2(n + 1)
D.None of these
Solution
We have
C02 - 2C12 + 3C22 - 4C32 + ...... +(-1)n(n + 1) Cn2
= {C02 - C12 + C22 - C32 + ..... +(-1)nCn2}
- {C12 - 2C22 + 3C32 - ..... + (-1)n nCn2}
= (-2)n/2
= (-1)n/2
∴
(-1)n/2 
{C02 - 2C12 + 3C22 - ....... + (-1)r (n + 1)Cn2}
= (-1)n/2 (n + 2)
C02 - 2C12 + 3C22 - 4C32 + ...... +(-1)n(n + 1) Cn2
= {C02 - C12 + C22 - C32 + ..... +(-1)nCn2}
- {C12 - 2C22 + 3C32 - ..... + (-1)n nCn2}
= (-2)n/2
= (-1)n/2
∴
(-1)n/2 {C02 - 2C12 + 3C22 - ....... + (-1)r (n + 1)Cn2} = (-1)n/2 (n + 2)
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