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Sum
Find the 4th term from the end in the expansion of `(x^3/2 - 2/x^2)^9`
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Solution
Since rth term from the end in the expansion of (a + b)n is (n – r + 2)th term from the beginning.
Therefore 4th term from the end is 9 – 4 + 2
i.e., 7th term from the beginning
Which is given by T7 = `""^9"C"_6 (x^3/2) ((-2)/x^2)^6`
= `""^9"C"_3 x^9/8 * 64/x^12`
= `(9 xx 8 xx 7)/(3 xx 2 xx 1) xx 64/x^3`
= `672/x^3`
Concept: Binomial Theorem for Positive Integral Indices
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