Q.5) Binomial Theorem

Q : Two consecutive terms in the expansion of (3 + 2x)^(74) whose coeff. are equal, are ________?



SOLUTION :  Let (r+1)th and (r+2)th terms are two consecutive terms in the expansion of (3+2x)^74.

By condition, T(r + 1) = T(r + 2)
and (r + 1)th term in the expansion of (a + x)^n is given by (nCr) * a^(n - r) * (x)^r

=> (74Cr) * 3^(74 - r) * (2x)^r = [74C(r+1)] * 3^(74 - r - 1) * (2x)^(r+1)
=> (74Cr) * 3^(74 - r) * (2)^r = [74C(r+1)] * 3^(74 - r - 1) * (2)^(r+1) - - {only coefficients}

=> (2^r) / (74 - r)! = 2^(r + 1) / 3 (r + 1) (73 - r)!
=> 1 / (74 - r) = 2 / [3 (r + 1)]
=> 3r + 3 = 148 - 2r
=> 5r = 145
=> r = 29

so, (r + 1)th and (r + 2)th terms are 30th and 31st terms. (Ans)