<p>The sum of odd degree terms in the expansion of \((x + \sqrt{x^3-1})^5 + (x - \sqrt{x^3-1})^5\) is:</p>
<p>\(10x^7 + 20x^5 + 2x^2 - 10x^4 - 20x^3 + 10x\)</p>
<p>\(2\)</p>
<p>\(0\)</p>
<p>\(2(5x^7 + 10x^6 + x^5 - 10x^4 - 10x^3 + 5x)\)</p>
Step-by-Step Solution
Key Concept: When two binomial expansions with opposite middle terms are added, all even-indexed terms cancel out due to alternating signs. You need to identify which terms in the final expansion have odd degree in x and sum only those.
<p><strong>Step 1:</strong> Let $y = \sqrt{x^3-1}$. Consider $(x+y)^5 + (x-y)^5$.</p><p><strong>Step 2:</strong> Using binomial theorem:</p><p>$(x+y)^5 = \sum_{r=0}^{5} \binom{5}{r}x^{5-r}y^r$</p><p>$(x-y)^5 = \sum_{r=0}^{5} \binom{5}{r}x^{5-r}(-y)^r$</p><p><strong>Step 3:</strong> Adding these expansions, terms with odd r cancel (since $(-1)^r = -1$), leaving only even r terms:</p><p>$(x+y)^5 + (x-y)^5 = 2\left[\binom{5}{0}x^5 + \binom{5}{2}x^3y^2 + \binom{5}{4}xy^4\right]$</p><p><strong>Step 4:</strong> Substitute $y^2 = x^3-1$:</p><p>$= 2\left[x^5 + 10x^3(x^3-1) + 5x(x^3-1)^2\right]$</p><p>$= 2\left[x^5 + 10x^6 - 10x^3 + 5x(x^6-2x^3+1)\right]$</p><p>$= 2\left[x^5 + 10x^6 - 10x^3 + 5x^7 - 10x^4 + 5x\right]$</p><p><strong>Step 5:</strong> Identify odd degree terms: $x^5, x^3, x$ and $x^7$ (odd powers).</p><p>Sum of odd degree terms: $2[5x^7 + 10x^6 + x^5 - 10x^4 - 10x^3 + 5x]$ where only odd powers contribute.</p><p>∴ Answer: <strong>B</strong></p>
Correct Answer: B