AP EAMCET201920 Apr 2019Evening ShiftMathematicsBinomial TheoremActual
If (1+x+x^2 )^n=c₀+c₁ x+c₂ x^2+ , then the value of c₀ c₁-c₁ c₂+c₂ c₃- is
Options
- A(-1)^n
- B0
- C2^n
- D3^n
Correct answer
B. 0
Step-by-step solution
On replacing x by - 1 x , we get aligned & (1-x+x^2 )^n x^ 2 n & = C₀ x^ 2 n -C₁ x^ 2 n-1 +C₂ x^ 2 n-2 -C₃ x^ 2 n-3 + x^ 2 n & (1-x+x^2 )^n=C₀ x^ 2 n -C₁ x^ 2 n-1 +C₂ x^ 2 n-2 aligned aligned & C₀ C₁-C₁ C₂+C₂ C₃- = Coefficient of x^ 2 n+1 in & (C₀ x^ 2 n -C₁ x^ 2 n-1 +C₂ x^ 2 n-2 -C₃ x^ 2 n-3 + ) & (C₀+C₁ x+C₂ x^2+C₃ x^3+ ) & = Coefficient of x^ 2 n+1 in (1+x+x^2 )^n (1-x+x^2 )^n & = Coefficient of x^ 2 n+1 in ( (1+x^2 )^2-x^2 )^n & = Coefficient of x^ 2 n+1 in (1+x^4+x^2 )^n=0 aligned Coefficient of x^k in the exp