Given $S=1+3+5+\cdots+2017+2019$, find $\frac{1}{1010}S-1008$.

Question

I'm wrong, or the answer in the book is wrong:

Given $S=1+3+5+\cdots+2017+2019$, find $$\frac{1}{1010}S-1008$$

My attempt is:

$S=1+3+5+\cdots+2017+2019$

$S=\frac{2019-1}{2}(1+2019)$

$S=2038180$

Now I find $$\frac{1}{1010}S-1008=\frac{1}{1010}2038180-1008=1010$$

But in the book answer is $2$. Where is my error, or is the answer in the book wrong? Help me please.

Answer 1

$$S_3:=1+3+5=9,\\\frac{S_3}3-(3-2)=2$$ and more generally

$$S_n=1+3+5+\cdots(2n-1)=n^2,\\\frac Sn-(n-2)=2.$$

Answer 2

You could solve it as follows:

Each term is odd, so it may be written as $2n+1$. Noticing that $2019=2\cdot 1009+1$ The sum is then

$$S=\sum_{n=0}^{1009}(2n+1)=2\sum_{n=0}^{1009}n+\sum_{n=0}^{1009}1$$

$$S=2\left (\frac{1009\cdot 1010}{2}\right)+1010=1010^2$$

Thus, $$\frac{S}{1010}-1008=2$$

Answer 3

The answer is actually 2. Assuming a=1, d=2;

  a+(n–1)d=2019

or, n–1=1009

>>n=1010

Now, S=n/2×{2a+(n–1)d}

S=1010/2×{2+(1009×2)}

S=1010^2

Thus, (1010^2/1010)–1008=2

So, the answer is 2