Question 195 of 246MediumMCQ
\[\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{r \over {2{r^2} - 7rn + 6{n^2}}}} \] is equal to :
Solutions · 1
Official solution
\[\mathop {\lim }\limits_{n \to \alpha } \sum\limits_{r = 1}^n {{r \over {2{r^2} - 7rn + 6{n^2}}}} \]
\[ = \mathop {\lim }\limits_{n \to \alpha } {1 \over n}\sum\limits_{r = 1}^n {{{\left( {{r \over n}} \right)} \over {2{{\left( {{r \over n}} \right)}^2} - 7\left( {{r \over n}} \right) + 6}}} \]
\[ = \int_a^b {f(x)dx} \]
\[a = \mathop {\lim }\limits_{n \to \alpha } \left( {{1 \over n}} \right) = 0\]
\[b = \mathop {\lim }\limits_{n \to \alpha } \left( {{n \over n}} \right) = 1\]
and \[{r \over n} \to x\]
\[ = \int_0^1 {{x \over {2{x^2} - 7x + 6}}dx} \]
\[ = \int_0^1 {{x \over {2{x^2} - 3x - 4x + 6}}dx} \]
\[ = \int_0^1 {{x \over {(2x - 3)(x - 2)}}dx} \]
\[ = \int_0^1 {\left[ {{A \over {(2x - 3)}} + {B \over {(x - 2)}}} \right]dx} \]
\[ = \int_0^1 {\left( {{{ - 3} \over {2x - 3}} + {2 \over {x - 2}}} \right)dx} \]
\[ = \left[ { - {{3\log |2x - 3|} \over 2} + 2\log |x - 2|} \right]_0^1\]
\[ = - {3 \over 2}\left[ {\log ( - 1) - \log ( - 3)} \right] + 2\left[ {\log ( - 1) - \log ( - 2)} \right]\]
\[ = - {3 \over 2}\log \left( {{1 \over 3}} \right) + 2\log \left( {{1 \over 2}} \right)\]
\[ = + {3 \over 2}\log 3 - 2\log 2\]
\[ = \log \sqrt {{3^3}} - \log 4\]
\[ = \log {{\sqrt {{3^2} \times 3} } \over 4}\]
\[ = \log {{3\sqrt 3 } \over 4}\]
Answer the question to unlock