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Question
Find the sum of the following serie:
(a − b)2 + (a2 + b2) + (a + b)2 + ... + [(a + b)2 + 6ab]
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Solution
(a − b)2 + (a2 + b2) + (a + b)2 + ... + [(a + b)2 + 6ab]
Here, the series is an A.P. where we have the following:
\[a = (a - b )^2 \]
\[d = \left( a^2 + b^2 - (a - b )^2 \right) = 2ab\]
\[ a_n = {[(a + b)}^2 + 6ab]\]
\[ \Rightarrow (a - b )^2 + (n - 1)(2ab) = {[(a + b)}^2 + 6ab] \]
\[ \Rightarrow a^2 + b^2 - 2ab + 2abn - 2ab = [ a^2 + b^2 + 2ab + 6ab]\]
\[ \Rightarrow a^2 + b^2 - 4ab + 2abn = a^2 + b^2 + 8ab\]
\[ \Rightarrow 2abn = 12ab \]
\[ \Rightarrow n = 6\]
\[ S_n = \frac{n}{2}\left[ 2a + (n - 1)d \right]\]
\[ \Rightarrow S_6 = \frac{6}{2}\left[ 2(a - b )^2 + \left( 6 - 1 \right) 2ab \right] \]
\[ = 3\left[ 2( a^2 + b^2 - 2ab) + 10ab \right]\]
\[ = 3\left[ 2 a^2 + 2 b^2 - 4ab + 10ab \right]\]
\[ = 3\left[ 2 a^2 + 2 b^2 + 6ab \right]\]
\[ = 6\left[ a^2 + b^2 + 3ab \right]\]
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