No. 111!

$\sum _{ y=1 }^{ 111 }{ \frac { { y }^{ 4 }+{ y }^{ 2 }+1 }{ { y }^{ 4 }+y } } \\$

Now, since, we know that: ${ a }^{ 3 }-{ b }^{ 3 }\quad =\quad (a-b)({ a }^{ 2 }+ab+{ b }^{ 2 })$

PUTTING

$a\quad =\quad { y }^{ 2 },\quad b\quad =\quad 1$

WE GET : $\quad \quad { y }^{ 6 }-1\quad =\quad ({ y }^{ 2 }-1)({ y }^{ 4 }+{ y }^{ 2 }+1)\\ \Rightarrow ({ y }^{ 4 }+{ y }^{ 2 }+1)\quad =\quad \frac { { y }^{ 6 }-1 }{ ({ y }^{ 2 }-1) }$ -------1.

ALSO

${ y }^{ 4 }+y\quad =\quad y({ y }^{ 3 }+1)$ -------2.

Using in the main equation, we need to find:

$\quad \quad \sum _{ y=1 }^{ 111 }{ \frac { \frac { { y }^{ 6 }-1 }{ ({ y }^{ 2 }-1) } }{ y({ y }^{ 3 }+1) } } \\ \\ \Rightarrow \sum _{ y=1 }^{ 111 }{ \frac { { y }^{ 6 }-1 }{ y({ y }^{ 3 }+1)({ y }^{ 2 }-1) } } \\ \\ \Rightarrow \sum _{ y=1 }^{ 111 }{ \frac { { (y }^{ 3 }-1)({ y }^{ 3 }+1) }{ y({ y }^{ 3 }+1)({ y }^{ 2 }-1) } } \\ \\ \Rightarrow \sum _{ y=1 }^{ 111 }{ \frac { { (y }^{ 3 }-1) }{ y({ y }^{ 2 }-1) } } \\ \\ \Rightarrow \sum _{ y=1 }^{ 111 }{ \frac { { (y }-1)({ y }^{ 2 }+y+1) }{ y({ y }-1)(y+1) } } \\ \\ \Rightarrow \sum _{ y=1 }^{ 111 }{ \frac { ({ y }^{ 2 }+y+1) }{ y(y+1) } } \\ \\ \Rightarrow \sum _{ y=1 }^{ 111 }{ \frac { { (y+1) }^{ 2 }-y }{ y(y+1) } } \\ \\ \Rightarrow \sum _{ y=1 }^{ 111 }{ \frac { y+1 }{ y } } -\sum _{ y=1 }^{ 111 }{ \frac { 1 }{ y+1 } } \\ \\ \Rightarrow \sum _{ y=1 }^{ 111 }{ 1 } +\sum _{ y=1 }^{ 111 }{ \frac { 1 }{ y } } -\sum _{ y=1 }^{ 111 }{ \frac { 1 }{ y+1 } }$

Now, since there is no fixed formula, we need to identify a pattern. See, by putting initial values, we can see that, the consecutive terms do get cancelled.

Since $\sum _{ y=1 }^{ 111 }{ 1 } =111$ , so the question is reduced to: $\quad \quad \quad 111+\sum _{ y=1 }^{ 111 }{ \frac { 1 }{ y } } -\sum _{ y=1 }^{ 111 }{ \frac { 1 }{ y+1 } } \\ \\ \Rightarrow \quad 111+1-\frac { 1 }{ 2 } +\frac { 1 }{ 2 } -\frac { 1 }{ 3 } +....+\frac { 1 }{ 111 } -\frac { 1 }{ 112 } \\ \Rightarrow \quad 111+1-\frac { 1 }{ 112 } \\ \Rightarrow \quad \frac { 12543 }{ 112 } \quad =\quad \frac { a }{ b } \\ \Rightarrow \quad a+b\quad =\quad 12655\\ \Rightarrow \quad (a+b)\quad \times \quad 111\quad =\quad 1404705\quad$

Phew! Cheers!!:):)

[n(n+2)/n+1 ]=[a/b] we have n=111 & n+2=113 & n+1=112
a=111 113= 12543 & b=112 so 111 (a+b)= 111*12655= 1404705

Let's factor the argument of the summation.

$\displaystyle \frac{y^4+y^2+1}{y^4+y}$

$\displaystyle = \frac{(y^2+y+1)(y^2-y+1)}{y(y+1)(y^2-y+1)}$ (See Sophie Germain identity and sum of cubes)

$\displaystyle = \frac{y^2+y+1}{y^2+y}$

$\displaystyle = 1 + \frac{1}{y^2+y}$

$\displaystyle = 1 + \left(\frac{1}{y} - \frac{1}{y+1}\right)$ (Partial fraction decomposition)

Now we can substitute this into the summation.

$\displaystyle \sum_{y=1}^{111} 1 + \frac{1}{y} - \frac{1}{y+1}$

$\displaystyle = 111 + \sum_{y=1}^{111} \frac{1}{y} - \frac{1}{y+1}$

$\displaystyle = 111 + 1 - \frac{1}{112}$ (Telescoping Series)

$\displaystyle = \frac{12543}{112}$

$a + b = 12655$ , then $111(a+b) = \boxed{1404705}$ .