Find the middle terms in the expansion of (5x−7y)7.
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Solution
Since n is an odd number, the expansion contains two middle terms. (7+12)th and (7+32)th terms are the two middle terms in the expansion of (5x−7y)7. T4=T3+1=(−1)37C3(5x)7−3(7y)3=−7C3(5x)4(7y)3 T5=T4+1=(−1)47C4(5x)7−4(7y)4=−7C4(5x)3(7y)4