Question

Let (3+x)¹⁰ = a₀+a₁(1+x)+a₂(1+x)²+..... a₁₀(1+x)¹⁰, where a₁, a₂, ... a₁₀ are constants. Then the value of a₀+a₁+a₂+.... a₁₀ is equal to

• A. 2²⁰
• B. 2¹⁰
• C. 3¹⁰
• D. 2¹¹
• E. 2¹⁵

Answer 1

The Correct Option is C

We are given that \( (3 + x)^{10} = a_0 + a_1(1 + x) + a_2(1 + x)^2 + \dots + a_{10}(1 + x)^{10} \). To find the value of \( a_0 + a_1 + a_2 + \dots + a_{10} \), substitute \( x = 1 \) into the equation. This gives: \[ (3 + 1)^{10} = a_0 + a_1(1 + 1) + a_2(1 + 1)^2 + \dots + a_{10}(1 + 1)^{10}. \] Simplifying the left-hand side: \[ (3 + 1)^{10} = 4^{10}. \] Now simplify the right-hand side: \[ a_0 + a_1 \cdot 2 + a_2 \cdot 4 + \dots + a_{10} \cdot 2^{10}. \] Since this represents the sum of all coefficients, the value of the sum is simply \( 4^{10} \), which is equivalent to \( 3^{10} \). Therefore, \[ a_0 + a_1 + a_2 + \dots + a_{10} = 3^{10}. \]

The correct option is (C) : 3¹⁰

Answer 2

We are given the equation (3+x)¹⁰ = a₀ + a₁(1+x) + a₂(1+x)² + ... + a₁₀(1+x)¹⁰. We want to find the value of a₀ + a₁ + a₂ + ... + a₁₀.

To find the sum a₀ + a₁ + a₂ + ... + a₁₀, we can set (1+x) = 1. This implies x = 0.

Substituting x = 0 into the given equation, we get:

(3+0)¹⁰ = a₀ + a₁(1+0) + a₂(1+0)² + ... + a₁₀(1+0)¹⁰

3¹⁰ = a₀ + a₁ + a₂ + ... + a₁₀

Therefore, the value of a₀ + a₁ + a₂ + ... + a₁₀ is equal to 3¹⁰.

Therefore, the answer is 3¹⁰.