Binomial Theorem
Odd and even terms
Grade 11
Question:
<p>In the expansion of \((x+a)^n\) if the sum of odd terms is \(P\) and the sum of even terms is \(Q\), then</p>
<p>(1) \(P^2 - Q^2 = (x^2 - a^2)^n\)</p>
<p>(2) \(4PQ = (x+a)^{2n} - (x-a)^{2n}\)</p>
<p>(3) \(2(P^2 + Q^2) = (x+a)^{2n} + (x-a)^{2n}\)</p>
<p>(4) none of these</p>
Step-by-Step Solution
Key Concept: In the binomial expansion of (x+a)^n, odd-positioned terms and even-positioned terms form distinct groups. Using the binomial theorem with substitution x→1 and x→-1 separately allows us to isolate these sums algebraically.
<p><strong>Step 1:</strong> Write the expansion: $(x+a)^n = \binom{n}{0}x^n + \binom{n}{1}x^{n-1}a + \binom{n}{2}x^{n-2}a^2 + ... + \binom{n}{n}a^n$</p><p><strong>Step 2:</strong> Sum of odd-positioned terms (1st, 3rd, 5th...): $P = \binom{n}{0}x^n + \binom{n}{2}x^{n-2}a^2 + \binom{n}{4}x^{n-4}a^4 + ...$</p><p><strong>Step 3:</strong> Sum of even-positioned terms (2nd, 4th, 6th...): $Q = \binom{n}{1}x^{n-1}a + \binom{n}{3}x^{n-3}a^3 + \binom{n}{5}x^{n-5}a^5 + ...$</p><p><strong>Step 4:</strong> Use the identity: $(x+a)^n = P + Q$ and $(x-a)^n = P - Q$</p><p><strong>Step 5:</strong> From these two equations: $P = \frac{(x+a)^n + (x-a)^n}{2}$ and $Q = \frac{(x+a)^n - (x-a)^n}{2}$</p><p><strong>Step 6:</strong> Key relations: $P^2 - Q^2 = (P+Q)(P-Q) = (x+a)^n \cdot (x-a)^n = (x^2-a^2)^n$ and $P + Q = (x+a)^n$; $P - Q = (x-a)^n$</p><p>∴ Answer: BC</p>
Correct Answer: BC