If Z[yn] = Y(2) then Z[Yn+3] :
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The given expression in for of sum can be written as
∑r=0n(−1)rxyzCr+∑r=0n(−1)rr(x+y+z)Cr+∑r=0n(−1)rr2(xy+yz+zx)Cr−∑r=0n(−1)rr3(xyz)Cr
Since n>3, the above terms form alternate positive and negative pairs thus cancelling each other and yielding zero.
Thus the final answer is 0.
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