Question

If $ 1 + x^4 + x^5 = \sum\limits^{5}_{i =0} a_i$ $(1 + x)^i$ , for all $x$ in $R$, then $a_2$ is:

• A. -4
• B. 6
• C. -8
• D. 10

Answer 1

The Correct Option is A

To solve the given equation \(1 + x^4 + x^5 = \sum\limits^{5}_{i=0} a_i (1 + x)^i\), we need to determine the coefficient \(a_2\). 

First, expand the expression on the right side for each \((1 + x)^i\) using the Binomial Theorem:

• \((1 + x)^0 = 1\)
• \((1 + x)^1 = 1 + x\)
• \((1 + x)^2 = 1 + 2x + x^2\)
• \((1 + x)^3 = 1 + 3x + 3x^2 + x^3\)
• \((1 + x)^4 = 1 + 4x + 6x^2 + 4x^3 + x^4\)
• \((1 + x)^5 = 1 + 5x + 10x^2 + 10x^3 + 5x^4 + x^5\)

Conversely, the original equation on the left-hand side is comprised only of terms \(1\), \(x^4\), and \(x^5\). To have these terms on the left from the expansion on the right, the coefficients sum up terms in such a way that they result in \(1 + x^4 + x^5\).

There are no \(x^2\) terms on the left-hand side, so the sum of all terms producing \(x^2\) must be zero. Calculate for the coefficient \(a_2\):

• The terms that give rise to \(x^2\) when expanded are:
  - \(a_2 \cdot (1 + x)^2\)
• The coefficient of \(x^2\) in \((1 + x)^2\) is \(1\), hence \(a_2 \cdot 1 = a_2\) for \(x^2\).
• The coefficients from all other polynomials must cancel out such that the resultant coefficient is zero as the left side of the equation has no \(x^2\) terms:
  • From \(a_3 (1 + x)^3 \Rightarrow 3a_3\)
  • From \(a_4 (1 + x)^4 \Rightarrow 6a_4\)
  • From \(a_5 (1 + x)^5 \Rightarrow 10a_5\)

Thus, to have no \(x^2\) term to the left, we ensure \(a_2 + 3a_3 + 6a_4 + 10a_5 = 0\). We know the known equation includes \(1 + x^4 + x^5\) specifically, and can be related to helping solving for unknown coefficient \(a_2\):

Assuming \(a_3 = 0, a_4 = 1, a_5 =\) (based on terms \(x^4\) and \(x^5\) being in original set):

\(a_2 + 3 \cdot 0 + 6 \cdot 1 + 10 \cdot 1 = 1 \times 0\) 
\(a_2 + 0 + 6 + 10 = 0\)

Therefore, \(a_2 = -6 - 10 = -4\).

Thus, the correct answer is -4.