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1 solution
We\quad can\quad write\quad this\quad as:\\ \Rightarrow \frac { 4(1) }{ 1.3 } +\frac { 4(2) }{ 3.7 } +\frac { 4(3) }{ 7.13 } +\frac { 4(4) }{ 13.21 } +\frac { 4(5) }{ 21.31 } +......\infty \\ let\quad S=1+3+7+13+21+....{ .......T. }_{ n }\quad \longrightarrow 1\\ \quad \quad S=\quad \quad 1+3+7+13+............+{ T }_{ n }\longrightarrow 2\\ by\quad 1-2\\ { T }_{ n }=1+(2+4+6+8+10.........(n-1))\\ { T }_{ n }={ n }^{ 2 }-n+1\\ \\ let\quad k=3+7+13+21+31+.........K_{ n }\longrightarrow 3\\ \quad \quad k=\quad \quad 3+7+13+21+..........+K_{ n }\longrightarrow 4\\ by\quad 3-4\\ K_{ n }=3+(4+6+8+10+.......(n-1))\\ K_{ n }={ n }^{ 2 }+n+1\\ \\ \therefore \quad General\quad Term\quad =\quad \frac { 4n }{ ({ n }^{ 2 }-n+1)({ n }^{ 2 }+n+1) } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad 2\left[ \frac { 1 }{ { n }^{ 2 }-n+1 } -\frac { 1 }{ { n }^{ 2 }+n+1 } \right] \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \quad 2\left[ \frac { 1 }{ 1 } -\frac { 1 }{ 3 } +\frac { 1 }{ 3 } -......... \right] \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad 2\\ \\ \quad \quad \quad \quad