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The number of real negative terms in the binomial expansion of `(1+i x)^(4n-2),n in N ,x >0`is`n`b. `n+1`c. `n-1`d. `2n`A. nB. n+1C. n-1D. 2n |
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Answer» Correct Answer - b Let `T_(r+1)` be the `(r +1)^(th)` term in the binomial expanion of `(1 + ix)^(4n -2)` . Then `T_(r+1) = ""^(4n-2)C_(r)(ix)^(r) = ""^(2n-2)C_(r)x^(r)i^(r)` Clearly, `T_(r +1) ` is real negative , if ` r=2 ,6,10,14,…` These values of r from an A.P. with term 2 and common difference 4. lies between 2 and 4n -2 `therefore 2 le 4k - 2 le 4n - 2 rArr 1 le k le n` Hence, there are n real negative terms. |
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