Question

If (1+x)^2010 = C_0 + C_1x+C_2x^2...., then the sum of the series C_2 + C_5 + C_8 is?

Answer 1

The coefficients C_k in the expansion of (1 + x)²⁰¹⁰ are given by

\(C_k = (\frac{2010}{k})\)

We need to find the sum C₂ + C₅ + C₈ + … + C₂₀₀₉. This series consists of coefficients where the indices are of the form 2 + 3k for k = 0, 1, 2, ….

To extract the coefficients where the index is of the form 2 + 3k, we can use the roots of unity filter. Let ω = e^2πi/3 be a primitive cube root of unity. The sum can be calculated using:

\(S = \frac{1}{3} ((1 + 1)^{2010} + (1 + \omega)^{2010} + (1 + \omega^2)^{2010})\)

For the first term:

(1 + 1)²⁰¹⁰ = 2²⁰¹⁰

For the second term:

\(1 + \omega = 2 cos(\frac{\pi}{3}) = 1\)

(1 + ω)²⁰¹⁰ = 1²⁰¹⁰

= 1

For the third term:

\(1 + \omega^2 = 2cos(-\frac{\pi}{3}) = 1\)

(1 + ω²)²⁰¹⁰ = 1²⁰¹⁰ 

= 1

Now substituting back into the equation for S:

\(S = \frac{1}{3} (2^{2010} + 1 + 1) = \frac{1}{3}(2^{2010} + 2) = \frac{2^{2010} + 2}{3}\)

Thus, the sum C₂ + C₅ + C₈ + … + C₂₀₀₉ is:

\(\frac{2^{2010} + 2}{3}\)