The sum of series .^(20)C0-^(20)C1+^(20)...

The sum of series `.^(20)C_0-^(20)C_1+^(20)C_2-^(20)C_3++^(20)C_10` is `1/2` `.^(20)C_10` b. `0` c. `.^(20)C_10` d. `-^(20)C_10`

A

`1/2 .^(20)C_(10)`

B

0

C

`.^(20)C_(10)`

D

`-.^(20)C_(10)`

Text Solution

Verified by Experts

The correct Answer is:

A

We know that
`(1-1)^(20)=.^(20)C_(0)-.^(20)C_(1)+.^(20)C_(2)-.^(20)C_(3)+"...."+.^(20)C_(10)-.^(20)C_(11)+.^(20)C_(12)-"....."+.^(20)C_(26)=0`
`2(.^(20)C_(0)-.^(20)C_(1)+.^(20)C_(2)-.^(20)C_(3)+"...."-.^(20)C_(9))+.^(20)C_(10) = 0`
[`:' .^(20)C_(20)=.^(20)C_(0),.^(20)C_(19)=.^(20)C_(3)`, etc.]
`rArr .^(20)C_(0) - .^(20)C_(1) + .^(20)C_(2) - .^(20)C_(3)+"......"-.^(20)C_(9)+.^(20)C_(10)`
`= -1/2.^(20)C_(10)+.^(20)C_(10)= 1/2.^(20)C_(10)`.

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