Problem
Let a and b be two nonzero real numbers. If the coefficient of x5 in the expansion of (ax2 + 70/(27bx))4 is equal to the coefficient of x-5 in the expansion of (ax – 1/bx2)7 then find the value of 2b.
Solution
We’ll write the general term of both expansions, figure out the coefficients of the respective powers, and equate them to find the value of b. Looks like the term involving a will get cancelled out.
The general term in the first expansion is 4Cra4 – r(x2)4 – r70r/(27bx)r.
The power of x is 8 – 3r. Equating this to 5 gives r = 1.
So, the coefficient of x5 is 4C1a4 – 1701/(27b)1 or 280a3/27b.
Similarly, the general term in the second expansion is 7Cr(ax)7 – r(-1/bx2)r.
The power of x in this term is 7 – 3r. Equating this to -5 gives r = 4.
So, the required coefficient is 7C4(ax)7 – 4(-1/bx2)4 or 35a3/b4.
Since these two coefficients are equal, we have 280a3/27b = 35a3/b4.
As expected, the nonzero term a3 gets cancelled out.
The remaining equation gives b3 = 27/8, which means b = 3/2, or 2b = 3.
Comments
A textbook problem. Boring – not worthy of the JEE Advanced. Instead of including such problems, the examiners should reduce the number of problems, but maintain a certain threshold of difficulty level.