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Let $\DeclareMathOperator{S}{sgn}
\S (x)= \begin{cases}
+1 & \text{if } x \geq 0 \\
-1 & \text{if } x < 0
\end{cases}$

What is the value of the following summation?

$$\sum_{i=0}^{50} \S \left ( (2i - 1) (2i - 3) \dots (2i - 99)  \right)$$

  1. $0$
  2. $-1$
  3. $+1$
  4. $25$
  5. $50$
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3 Answers

Best answer
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1 votes
$\sum_{i=0}^{50}sgn((2i-1)(2i-3)\ldots(2i-99))$

There are 50(even number) terms present in the product.

Now we need to open summation for $i = 0$ to $50$

$i=0 , sgn((-1)(-3)\ldots(-99)) = +1$

Because we know that there are $50$ terms present in product which is an even number so it will be $+1.$

$i=1, sgn((1)(-1)\ldots (-97)) = -1$

Because in product there are $49$ negative terms and $1$ term is positive, answer will be negative.

From $i=0$ and $i=1,$ we understood that summation is giving $+1$ for even values of $i$ and $-1$ for odd values of $i.$

From $0$ to $50$ there are $26$ even and $25$ odd numbers are present.

So, answer will be $+1.$
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Given,
$$\S (x)= \begin{cases} +1 & \text{if } x \geq 0 \\ -1 & \text{if } x < 0 \end{cases}$$

We need to find,
$$\sum_{i=0}^{50} \S \left ( (2i - 1) (2i - 3) \dots (2i - 99)  \right)$$

$$
\begin{array}{|c|c|c|c|c|}
\hline
i & & \#-ve & \#+ve & \S & \text{Comments}\\
\hline
0 & \boxed{\color{red}{(-1)(-3)\dots(-99)}} & \color{red}{50} & \color{green}{0} & \color{green}{+1} & \text{#even of -ve int}\\
\hline
1 & \boxed{\color{green}{(1)}}\boxed{\color{red}{(-1)(-3)\dots(-97)}} & \color{red}{49} & \color{green}{1} & \color{red}{-1} & \text{#odd of -ve int}\\
\hline
2 & \boxed{\color{green}{(3)(1)}}\boxed{\color{red}{(-1)(-3)\dots(-95)}} & \color{red}{48} & \color{green}{2} & \color{green}{+1} & \text{#even of -ve int}\\
\hline
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots\\
\hline
k & \boxed{\color{green}{(2k-1)\dots(1)}}\boxed{\color{red}{(-1)\dots(2k-97)}} & \color{red}{50-k} & \color{green}{k} & \color{green}{} & \text{}\\
\hline
\end{array}
$$

We can observe that for even value of $i$, $\S$ would be $+1$ else $-1$. Hence we can get the answer by finding the difference of number of even values of $i$ and number of odd values of $i$.

$$
\begin{align}
\text{Number of even values of }i = 26 \\
\text{Number of odd values of }i = 25\\
\textbf{Answer} = 26.(+1) + 25.(-1) = \textbf{+1}
\end{align}
$$
 

$\textbf{Option (C) is correct}$
Answer:

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