Let the sixth term in the binomial expansion of ( 2^ 2(10 - 3^x) + 5 2^(x-2) 2 3)^m, in the increasing powers of 2^(x-2)

Question

Let the sixth term in the binomial expansion of ( 2^ 2(10 - 3^x) + 5 2^(x-2) 2 3)^m, in the increasing powers of 2^(x-2) 2 3, be 21. If the binomial coefficients of the second, third, and fourth…

Accepted Answer

Given: The sixth term of the expansion is $21$ . Also, the binomial coefficients of the second, third, and fourth terms are in the condition of an A.P.

Find: The sum of the squares of all possible values of $x$ .

From the solution, the sixth term is taken as

T_6 = \binom{m}{5}(10 - 3x)^{m-5} \cdot (3x^{-2})^5 = 21

Also,

\binom{m}{1}, \binom{m}{2}, \binom{m}{3}

are in arithmetic progression.

Using the A.P. condition,

2\binom{m}{2} = \binom{m}{1} + \binom{m}{3}

Substituting the binomial coefficient formulas,

2 \cdot \frac{m(m-1)}{2} = m + \frac{m(m-1)(m-2)}{6}

m(m-1) = 2m + \frac{m(m-1)(m-2)}{3}

3m(m-1) = 6m + m(m-1)(m-2)

m(m-1)(m-2) - 3m(m-1) + 6m = 0

m(m-1)(m-5) = 0

Hence, the accepted value is

m = 7

Now substitute $m = 7$ into the sixth-term condition:

\binom{7}{5} \cdot (10 - 3x)^2 \cdot \frac{(3x^{-2})^5}{9} = 21

Since

\binom{7}{5} = 21

we get

21 \cdot (10 - 3x)^2 \cdot \frac{3^5}{9x^{10}} = 21

(10 - 3x)^2 \cdot \frac{3^5}{9x^{10}} = 1

Let

y = 3x^{-2}

Then the equation becomes

10y - y^2 = 1

which gives

y^2 - 10y + 1 = 0

Using the quadratic formula,

y = \frac{10 \pm \sqrt{100 - 4}}{2} = \frac{10 \pm \sqrt{96}}{2} = \frac{10 \pm 4\sqrt{6}}{2}

So,

y = 5 \pm 2\sqrt{6}

Now,

3x^{-2} = 5 + 2\sqrt{6} \quad \text{or} \quad 3x^{-2} = 5 - 2\sqrt{6}

From the extracted solution, the sum of the squares of all possible values of $x$ is

4

Therefore, the required answer is $4$ .

Given: The condition on the second, third, and fourth term coefficients leads to an arithmetic progression.

Find: First determine $m$ , then use the sixth term condition.

For the binomial expansion of $(a+b)^m$ , the coefficients of the second, third, and fourth terms are

\binom{m}{1}, \binom{m}{2}, \binom{m}{3}

If they are in A.P., then

2\binom{m}{2} = \binom{m}{1} + \binom{m}{3}

Substitute:

2 \cdot \frac{m(m-1)}{2} = m + \frac{m(m-1)(m-2)}{6}

After simplification, the extracted solution gives the admissible value

m = 7

Then the sixth term equation is solved and reduced by the substitution

y = 3x^{-2}

This leads to

y^2 - 10y + 1 = 0

with roots

y = 5 \pm 2\sqrt{6}

Using $y = 3x^{-2}$ , the final result reported in the solution is that the sum of the squares of all possible values of $x$ equals $4$ .

Therefore, the answer is $4$ .

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