TIFR CSE 2013 | Part A | Question: 15

Question

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?

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

• A. $0$
• B. $-1$
• C. $+1$
• D. $25$
• E. $50$

Comments

@srestha Ma'am 

I think answer should be -1 .

there are 45 terms in this sequencsequence.So sgn(0) will be -1 .  likewise . even will give -1 and odd will give +1.

now, 0 and 1 will make pair and result in 0.

so forth... 8 and 9 will be paired and result in 0.

48 and 49 will make pair and result in 0.

so we left with 50 which is even so it will give "-1" .

 

p.s : yes key is +1 but where i am making mistake then ?

---

@Headshot, for even it will be +1 and for odd it will be -1 check answer below.

Answer 1

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

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 $+1.$

Comments

@Shubhgupta 

Please check : 

first term : 1 

last term : 99

difference : 2

$\frac{99-1}{2} + 1$

so there are 45 terms which is an odd number.

i = 0 to 50 , is "i" value , why you considered it as number of terms ?

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@HeadShot,

$\small \frac{99-1}{2} +1 = 45 How?$ should b 50 right. $\small \frac{98}{2}+1 = 49 +1 = 50.$

For second doubt check answer i was not considering i as no. of terms. I have framed correctly now.

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@Shubhgupta 

Oh..my bad.. calculated no. of terms wrong. 

thanx.

Answer 2

s

so answer will be +1

Answer 3

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 $i$.

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

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