Problem With Sum

Question

I would like help with the following summation. Every time I come to this step in my signals class, I have difficulty proving that:

$$ \sum_{j=0}^{n-3}1=n-2 $$

Answer 1

Simply note that

$$\sum_{j=0}^{n-3}=\stackrel{n-2 \, times}{1=1+1+...+1}=n-2$$

Answer 2

The pattern is simple:

Let $S_n:=\sum_{j=0}^{n-3}1$. We have

$$S_3=1_0=1,\\S_4=1_0+1_1=2,\\S_5=1_0+1_1+1_2=3,\\S_6=1_0+1_1+1_2+1_3=4,\\\cdots$$

(the indices are there to explicit the values of $j$).

Answer 3

Increment the sum borders: $$\sum_{j=0}^{n-3} 1 = \sum_{j=0+1}^{n-3+1} 1 = \sum_{j=1}^{n-2} 1 = $$ $$\underbrace{1+1+\cdots+1}_{n-2}=n-2.$$

Answer 4

$$\sum_{j=0}^{n-3}1=n-2$$ Consider $\sum_{j=0}^{n-3}a_nx$ $$\sum_{j=0}^{n-3}a_jx=a_0x+\underbrace{a_1x+...a_{n-3}x}_{(n-3) terms } $$ Set $x=1$ and $a_j=1 $ for $j=0...$

So $(n-3)\times 1 +1=n-2$

Answer 5

A basic summation identity is: $$\sum_{j=1}^Nc=cN$$ With an extra term: $$\sum_{j=0}^Nc=c(1+N)$$