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Middle term in the expansion of `(a^(3) + ba)^(28)` is ...... |
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Answer» Given expansion is `(a^(3) + ba)^(28)` `:. N = 28` `:.` Middle term `= ((28)/(2) + 1)th` term = 15 th term `:. T_(15) = T_(14 + 1)` `= .^(28)C_(14) (a^(3))^(28 - 14) (ba)^(14)` `= .^(28)C_(14) a^(42) b^(14) a^(14)` `= .^(28)C_(14) a^(56) a^(14)` |
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