Let the sixth term in the binomial expansion of ( √2^(log ₂( 10 - 3ˣ )) + ⁵√2^((x - 2)log ₂3) )ᵐ in the increasing powers of 2^((x-2) log ₂ 3), be 21 . If the binomial coefficients of the second, third and fourth terms in the expansion are respectively the first, third and fifth terms of an A.P., then the sum of the squares of all possible values of x is __________.

Solution

Maths sir · 5 steps

Get m from the A.P. condition first. Then the sixth term becomes a quadratic in 3ˣ.

1. Clean up the terms. Since 2^(log₂ a) = a, the first term is √(10-3ˣ) = (10-3ˣ)^(1/2). The second is (2^(log₂ 3ˣ⁻²))^(1/5) = 3^((x-2)/5).
2. Find m. The 1st, 3rd and 5th terms of an A.P. are a, a+2d, a+4d. So the middle one is the average of the other two: 2 ᵐC₂ = ᵐC₁ + ᵐC₃. This simplifies to m² - 9m + 14 = 0, so m = 2 or 7. A sixth term needs m ≥ 5, so m = 7.
3. Sixth term. In increasing powers of the second term, Tᵣ₊₁ = ⁷Cᵣ (10-3ˣ)^((7-r)/2) (3^((x-2)/5))ʳ. Take r = 5 and set the term equal to 21.

   ⁷C₅ (10-3ˣ) 3ˣ⁻² = 21 ⇒ (10-3ˣ) 3ˣ/9 = 1

4. Solve for x. Put t = 3ˣ. Then t² - 10t + 9 = 0, so t = 1 or t = 9. That gives x = 0 or x = 2. Both keep 10 - 3ˣ > 0, so both are valid.
5. So. The sum of the squares is 0² + 2² = 4.

The common slip: 2. Adding the values of x instead of their squares gives 0 + 2 = 2, not 4.

The idea to keep. When terms look like 2^(log₂ a), collapse them to a first. Use the A.P. condition to fix n, then write the general term Tᵣ₊₁ and set it equal to the given value.

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