AP EAMCET2008MathematicsBinomial Theorem
If (1+x+x^2+x^3 )^5= _ k=0 ¹⁵ a_k x^k , then _ k=0 ^7 a_ 2 k is equal to
Options
- A128
- B256
- C512
- D1024
Correct answer
C. 512
Step-by-step solution
aligned & Given, (1+x+x^2+x^3 )^5= _ k=0 ¹⁵ a_k x^k & [(1+x)+x(1+x)]^5= _ k=0 ¹⁵ a_k x^k & (1+x)¹⁰=a₀ x^0+a₁ x+a₂ x^2+ +a₁₅ x¹⁵ & ¹⁰ C₀+ ¹⁰ C₁ x+ ¹⁰ C₂ x^2+ + 10 10 x¹⁰ & =a₀+a₁ x+a₂ x^2+a₃ x^3+ +a₁₅ x¹⁵ aligned On equating the coefficient of constant and even powers of x , we get aligned a₀ & = ¹⁰ C₀, a₂= ¹⁰ C₂, a₄ & = ¹⁰ C₄, , a₁₀= 10 10 , a₁₂ & =a₁₄=0 _ k=0 ^7 a₂ k & = ¹⁰ C₀+ ¹⁰ C₂+ ¹⁰ C₄+ ¹⁰ C₆ & + ¹⁰ C₈+ 10 10 +0+0 & =2¹⁰⁻¹=2^9 & =512 aligned