Let the sum of the following infinite sequence be $N$ . $1,\dfrac{1}{2},\dfrac{1}{2},\dfrac{1}{6},\dfrac{1}{4},\dfrac{1}{12},\dfrac{1}{8},\dfrac{1}{20},\dfrac{1}{16},\dfrac{1}{30},\dfrac{1}{32},\dfrac{1}{42} \ldots \ .$ Find the $\textbf{remainder}$ when $\displaystyle \Bigl(N^{2014}+1\Bigr)$ is divided by $\color{#69047E}{\textbf{11}}$ .

**
Hint
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:- Is it really
$\textbf{a}$
sequence?

The answer is 5.

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See that the sequence is actually inter-twined mixing of $2$ sequences $\color{#3D99F6}{1},\color{#624F41}{\dfrac{1}{2}},\color{#3D99F6}{\dfrac{1}{2}},\color{#624F41}{\dfrac{1}{6}},\color{#3D99F6}{\dfrac{1}{4}},\color{#624F41}{\dfrac{1}{12}},\color{#3D99F6}{\dfrac{1}{8}},\color{#624F41}{\dfrac{1}{20}},\color{#3D99F6}{\dfrac{1}{16}},\color{#624F41}{\dfrac{1}{30}},\color{#3D99F6}{\dfrac{1}{32}},\color{#624F41}{\dfrac{1}{42}}....$

Out of these, the $\color{#3D99F6}{\textbf{sequence in Blue}}$ is $\displaystyle S_\infty= \color{#3D99F6}{1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8} }...$ which is a Geometric progression (GP) with $a=1,r=\dfrac{1}{2}$ , so it's infinite sum is

$\bullet \quad \dfrac{a}{1-r}\\ = \dfrac{1}{1-\frac{1}{2}}\\ = \dfrac{1}{\frac{1}{2}}\\ =\mathbf{2}$

The $\color{#624F41}{\textbf{sequence in Brown}}$ is $\color{#624F41}{\dfrac{1}{1\times 2} + \dfrac{1}{2\times 3}+\dfrac{1}{3\times 4}+\dfrac{1}{4\times 5}}.....$

This can be written as

$\dfrac{2-1}{1\times 2}+\dfrac{3-2}{2\times 3}+\dfrac{4-3}{3\times 4}+\dfrac{5-4}{4\times 5}+....$

$= \biggl(1-\dfrac{1}{2}\biggr)+\biggl(\dfrac{1}{2}-\dfrac{1}{3}\biggr)+\biggl(\dfrac{1}{3}-\dfrac{1}{4}\biggr)+\biggl(\dfrac{1}{4}-\dfrac{1}{5}\biggr)+....\quad \text{A telescoping series}\\ =\quad \mathbf{1}$

This gives $N=3$ .

We want to find $3^{2014}+1 \pmod{11}$

Observe that

$3^5 = 243\equiv 1 \pmod{11} \implies 3^{5k} \equiv 1 \pmod{11} \\ \therefore 3^{2010} \equiv 1 \pmod{11} \\ \therefore 3^{2014} \equiv 3^{4} \equiv 81 \equiv{4} \pmod{11}$

Thus, asked answer will be $3^{2014}+1\equiv 4+1 \equiv \boxed{5} \pmod{11}$