<p>If <span class='math'>(1 + x + x^2 + x^3)^5 = a_0 + a_1 x + a_2 x^2 + \cdots + a_{15} x^{15}</span>, then <span class='math'>a_{10}</span> equals</p>
<p>(a) <span class='math'>99</span></p>
<p>(b) <span class='math'>100</span></p>
<p>(c) <span class='math'>101</span></p>
<p>(d) <span class='math'>110</span></p>
Step-by-Step Solution
Key Concept: We need to find the coefficient of x^10 in (1 + x + x^2 + x^3)^5. This requires using the multinomial theorem or expressing the base as a geometric series and then expanding the fifth power systematically.
<p><strong>Step 1:</strong> Rewrite the expression using geometric series.
(1 + x + x^2 + x^3) = (1 - x^4)/(1 - x)</p><p>So (1 + x + x^2 + x^3)^5 = (1 - x^4)^5/(1 - x)^5</p><p><strong>Step 2:</strong> Expand (1 - x^4)^5 using binomial theorem:
(1 - x^4)^5 = Σ C(5,k)(-1)^k x^(4k) for k = 0,1,2,3,4,5
= 1 - 5x^4 + 10x^8 - 10x^12 + 5x^16 - x^20</p><p><strong>Step 3:</strong> Expand (1 - x)^(-5) using the generalized binomial theorem:
(1 - x)^(-5) = Σ C(n+4,4)x^n for n = 0,1,2,...
where the coefficient of x^n is C(n+4,4)</p><p><strong>Step 4:</strong> Find coefficient of x^10 in the product (1 - 5x^4 + 10x^8 - 10x^12 + ...)(1/(1-x)^5).
The x^10 term comes from:
- 1 · [coefficient of x^10 in (1-x)^(-5)] = C(14,4)
- (-5x^4) · [coefficient of x^6 in (1-x)^(-5)] = -5·C(10,4)
- (10x^8) · [coefficient of x^2 in (1-x)^(-5)] = 10·C(6,4)</p><p><strong>Step 5:</strong> Calculate:
C(14,4) = 14·13·12·11/(4·3·2·1) = 1001
C(10,4) = 10·9·8·7/(4·3·2·1) = 210
C(6,4) = 6·5/(2·1) = 15</p><p><strong>Step 6:</strong> Coefficient of x^10:
a_10 = 1001 - 5(210) + 10(15)
= 1001 - 1050 + 150
= 101</p><p><strong>∴ Answer:</strong> C</p>
Correct Answer: C