The constant term in the expansion of ( 2x + 1/x⁷ + 3x² )⁵ is ___________.

Solution

Maths sir · 3 steps

Pick how many times each of the three terms is used, then make the powers of x add up to zero.

1. General term. Suppose 2x is chosen a times, 1/x⁷ is chosen b times and 3x² is chosen c times, with a + b + c = 5. By the multinomial theorem, that term is 5!/(a! b! c!)(2x)ᵃ(x⁻⁷)ᵇ(3x²)ᶜ.
2. Make power zero. The power of x is a - 7b + 2c, and it must be 0. If b = 0, then a + 2c = 0 forces a = c = 0, which breaks a + b + c = 5. If b ≥ 2, then a + 2c must be at least 14, but a + 2c ≤ 2(a + c) ≤ 6. So b = 1, and then a + c = 4 with a + 2c = 7 give c = 3, a = 1.

   5!/(1! 1! 3!) (2)¹ (1)¹ (3)³ = 20 × 2 × 27

3. So. The only constant term is 20 × 54 = 1080.

The common slip: 120. Using 3 instead of 3³ = 27 for the coefficient of 3x² gives 20 × 2 × 3 = 120, not 1080. The number in front gets raised to the same power as x².

The idea to keep. In a multinomial, write the power of x in terms of how often each term is chosen. Set it to the required power, then list the small whole-number cases one by one.

Answer from the NTA final answer key. Solution by PrepWiser.