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In the binomial expansion of `(a-b)^n , ngeq5,`the sum of 5thand 6th terms is zero, then `a/b`equals(1) `5/(n-4)`(2) `6/(n-5)`(3) `(n-5)/6`(4) `(n-4)/5`A. `(n - 5)/(6)`B. `(n - 4)/(5)`C. `(5)/(n - 4)`D. `(6)/(n - 5)` |
Answer» Given `T_5+T_6=0` `rArr .^Nc_4a^(n-4)b^4-.^nC_5a^(n-5)b^5=0` `rArr .^nC_4 a^(n-4)b^4=.^Nc_5 a^(n-5)b^5rArr (a)/(b)=(.^Nc_5)/(.^C_4)=(n-4)/(5)` |
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