Question
Show that $\frac { 1 \times 2 ^ { 2 } + 2 \times 3 ^ { 2 } + \ldots \ldots + n \times ( n + 1 ) ^ { 2 } } { 1 ^ { 2 } \times 2 + 2 ^ { 2 } \times 3 + \ldots \ldots + n ^ { 2 } ( n + 1 ) } = \frac { 3 n + 5 } { 3 n + 1 }$
$= \frac { \sum n ( n + 1 ) ^ { 2 } } { \sum n ^ { 2 } ( n + 1 ) } = \frac { \sum n \left( n ^ { 2 } + 2 n + 1 \right) } { \sum \left( n ^ { 3 } + n ^ { 2 } \right) }$
$= \frac { \sum n ^ { 3 } + 2 \sum n ^ { 2 } + \sum n } { \sum n ^ { 3 } + \sum n ^ { 2 } }$
$=\frac { \frac { n ^ { 2 } ( n + 1 ) ^ { 2 } } { 4 } + \frac { 2 n ( n + 1 ) ( 2 n + 1 ) } { 6 } + \frac { n ( n + 1 ) } { 2 } } { \frac { n ^ { 2 } ( n + 1 ) ^ { 2 } } { 4 } + \frac { n ( n + 1 ) ( 2 n + 1 ) } { 6 } }$
$=\frac { \frac { n ( n + 1 ) } { 2 } \left[ \frac { n ( n + 1 ) } { 2 } + \frac { 2 ( 2 n + 1 ) } { 3 } + 1 \right] } { \frac { n ( n + 1 ) } { 2 } \left[ \frac { n ( n + 1 ) } { 2 } + \frac { ( 2 n + 1 ) } { 3 } \right] }$
$= \frac { 3 n ^ { 2 } + 11 n + 10 } { 3 n ^ { 2 } + 7 n + 2 } = \frac { ( n + 2 ) ( 3 n + 5 ) } { ( n + 2 ) ( 3 n + 1 ) } = \frac { 3 n + 5 } { 3 n + 1 }$
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