Question

Let (x + 10)⁵⁰ + (x - 10)⁵⁰ = a₀ + a₁ x + a₂ x² +…+ a₅₀ x⁵⁰, for all x ∈ R; then \(\frac{{{a_2}}}{{{a_0}}}\) is equal to:
1. 12.50
2. 12.00
3. 12.25
4. 12.75

Answer 1

Correct Answer - Option 3 : 12.25

From question, the expression given is:

(x + 10)⁵⁰ + (x - 10)⁵⁰ = a₀ + a₁ x + a₂ x² +…+ a₅₀ x⁵⁰

Now,

⇒ (x + 10)⁵⁰ = ⁵⁰C₀ x⁵⁰ + ⁵⁰C₁ x⁴⁹ (10) +⁵⁰C₂ x⁴⁸ (10)²+…+⁵⁰C₅₀ (10)⁵⁰

⇒ (x - 10)⁵⁰ = ⁵⁰C₀ x⁵⁰ - ⁵⁰C₁ x⁴⁹ (10) + ⁵⁰C₂ x⁴⁸(10)^2-…+ ⁵⁰C₅₀ (10)⁵⁰

Now,

\(\Rightarrow \left[ {{}_\;^{50}{C_0}{x^{50}} + {}_\;^{50}{C_1}{x^{49}}\left( {10} \right) + {}_\;^{50}{C_2}{x^{48}}{{\left( {10} \right)}^2} + \ldots + {}_\;^{50}{C_{50}}{{\left( {10} \right)}^{50}}} \right] + \left[ {{}_\;^{50}{C_0}{x^{50}} - {}_\;^{50}{C_1}{x^{49}}\left( {10} \right) + {}_\;^{50}{C_2}{x^{48}}{{\left( {10} \right)}^2} - \ldots + {}_\;^{50}{C_{50}}{{\left( {10} \right)}^{50}}} \right] = {a_0} + {a_1}x + {a_2}{x^2} + \ldots + {a_{50}}{x^{50}}\)

\(\Rightarrow 2\left[ {{}_\;^{50}{C_0}{x^{50}} + {}_\;^{50}{C_2}{x^{48}}{{\left( {10} \right)}^2} + {}_\;^{50}{C_4}{x^{46}}{{\left( {10} \right)}^4} + \ldots + {}_\;^{50}{C_{48}}{x^2}{{\left( {10} \right)}^{48}} + {}_\;^{50}{C_{50}}{{\left( {10} \right)}^{50}}} \right] = {a_0} + {a_1}x + {a_2}{x^2} + \ldots + {a_{50}}{x^{50}}\)

Now, we need to compare the coefficient of a₂ and a₀.

Now,

⇒ a₀ = 2[⁵⁰C₅₀ (10)⁵⁰]

⇒ a₂ = 2[⁵⁰C₄₈ (10)⁴⁸]

From question,

\(\Rightarrow \frac{{{a_2}}}{{{a_0}}} = \frac{{2\left[ {{}_\;^{50}{C_{48}}{{\left( {10} \right)}^{48}}} \right]}}{{2\left[ {{}_\;^{50}{C_{50}}{{\left( {10} \right)}^{50}}} \right]}}\)

∵ [⁵⁰C₄₈ = ⁵⁰C₂]

\(\Rightarrow \frac{{{a_2}}}{{{a_0}}} = \frac{{2\left[ {{}_\;^{50}{C_2}{{\left( {10} \right)}^{48}}} \right]}}{{2\left[ {{}_\;^{50}{C_{50}}{{\left( {10} \right)}^{50}}} \right]}}\)

\(\Rightarrow \frac{{{a_2}}}{{{a_0}}} = \frac{{2\left[ {\left( {\frac{{50 \times 49}}{{1 \times 2}}} \right){{\left( {10} \right)}^{48}}} \right]}}{{2{{\left( {10} \right)}^{50}}}}\)

\(\Rightarrow \frac{{{a_2}}}{{{a_0}}} = \frac{{50 \times 49}}{{2\left( {10 \times 10} \right)}}\)

\(\Rightarrow \frac{{{a_2}}}{{{a_0}}} = \frac{{5 \times 49}}{{20}}\)

\(\Rightarrow \frac{{{a_2}}}{{{a_0}}} = \frac{{245}}{{20}}\)

\(\therefore \frac{{{a_2}}}{{{a_0}}} = 12.25\)