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Find the value of ^4n C0+^(4n)C4+^(4n)C8...

Find the value of `^4n C_0+^(4n)C_4+^(4n)C_8++""^(4n)C_(4n)` .

Text Solution

Verified by Experts

The correct Answer is:
`2^(4n-2)+(-1)^(n)2^(2n-1)`
We have,
`.^(4n)C_(0) + .^(4n)C_(2)x^(2)+.^(4n)C_(4)x^(4)+"…."+.^(4n)C_(4n)x^(4n)`
`= 1/2[(1+x)^(4n) +(1-x)^(4n)]`
Putting `x = 1` and `x= i`, we get
`.^(4n)C_(0) + .^(4n)C_(2) + .^(4n)C_(4) + "……" + .^(4n)C_(4n)=1/2[2^(4n)]`
and `.^(4n)C_(0) - .^(4n)C_(2) + .^(4n)C_(4) - "....." + .^(4n)C_(4n)`
`= 1/2[(1+i)^(4n)+(1-i)^(4n)]`
Thus, `2[.^(4n)C_(0) + .^(4n)C_(4)+"......"+.^(4n)C_(4n)] = 2^(4n-1) + 1/2[sqrt(2)(cos'(pi)/(4)- isin'(pi)/(4))]^(4n)`
`= 2^(2n) (cosnpi + isin npi)+2^(2n) (cos npi - i sin npi)`
`= 2^(2n+1) cos n pi=2^(2n+1)(-1)^(n)`
`:. 2[.^(4n)C_(0) + .^(4n)C_(4) + "....." + .^(4n)C_(4n)] = 2^(4n-1) + 1/2 2^(2n+1)(-1)^(n)`
`rArr .^(4n)C_(0) + .^(4n)C_(4) + "....." + .^(4n)C_(4n) = 2^(4n-2) + (-1)^(n)2^(2n-1)`
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