Question

Find the sum of the following serie:

(a − b)² + (a² + b²) + (a + b)² + ... + [(a + b)² + 6ab]

Answer 1

(a − b)² + (a² + b²) + (a + b)² + ... + [(a + b)² + 6ab]
Here, the series is an A.P. where we have the following:

\[a = (a - b )^2 \]

\[d = \left( a^2 + b^2 - (a - b )^2 \right) = 2ab\]

\[ a_n = {[(a + b)}^2 + 6ab]\]

\[ \Rightarrow (a - b )^2 + (n - 1)(2ab) = {[(a + b)}^2 + 6ab] \]

\[ \Rightarrow a^2 + b^2 - 2ab + 2abn - 2ab = [ a^2 + b^2 + 2ab + 6ab]\]

\[ \Rightarrow a^2 + b^2 - 4ab + 2abn = a^2 + b^2 + 8ab\]

\[ \Rightarrow 2abn = 12ab \]

\[ \Rightarrow n = 6\]

\[ S_n = \frac{n}{2}\left[ 2a + (n - 1)d \right]\]

\[ \Rightarrow S_6 = \frac{6}{2}\left[ 2(a - b )^2 + \left( 6 - 1 \right) 2ab \right] \]

\[ = 3\left[ 2( a^2 + b^2 - 2ab) + 10ab \right]\]

\[ = 3\left[ 2 a^2 + 2 b^2 - 4ab + 10ab \right]\]

\[ = 3\left[ 2 a^2 + 2 b^2 + 6ab \right]\]

\[ = 6\left[ a^2 + b^2 + 3ab \right]\]