Question

What is the value of $0-1+2-3+4-5+6-7.........+16-17+18-19+20$

Answer 1

Step 1. Split the given series.

The given series is $0-1+2-3+4-5+6-7.........+16-17+18-19+20$

Given series can be written as:

$$\begin{gathered} S=\left(0+2+4+6+8+....+18+20\right)-(1+3+5+....+17+19) \\ S=S_1-S_2 \end{gathered}$$

$$\begin{gathered} S_1=0+2+4+6+8+....+18+20 \\ {S}_2=1+3+5+....+17+19 \end{gathered}$$

Step 2. Calculate the sum of series $S_1$.

$S_1=0+2+4+6+8+....+18+20$

The common difference :

$$\begin{gathered} d=a_2-a_1 \\ =2-0 \\ =2 \end{gathered}$$

First-term, $a_1=0$

The general term expression of an AP is given as:

$$\begin{gathered} a_n=a_1+(n-1)d \\ \Rightarrow 20=0+(n-1)2 \\ \Rightarrow n=\frac{20}{2}+1 \\ \Rightarrow n=11 \end{gathered}$$

The summation of the first series is computed as follow:

$$\begin{gathered} S_1=\frac{n}{2}\left[2a_1+(n-1)d\right] \\ {S}_1=\frac{11}{2}\left[0+20\right] \\ {S}_1=110 \end{gathered}$$

Step 3. Calculate the sum of series $S_2$.

$S_2=1+3+5+....+17+19$

The common difference

$$\begin{gathered} d=a_2-a_1 \\ =3-1 \\ =2 \end{gathered}$$

First-term, $a_1=1$

The general term expression of an AP is given as follow:

$$\begin{gathered} a_n=a_1+(n-1)d \\ \Rightarrow 19=1+(n-1)2 \\ \Rightarrow n=\frac{18}{2}+1 \\ \Rightarrow n=10 \end{gathered}$$

The summation of the first series is computed as follow:

$$\begin{gathered} S_2=\frac{n}{2}\left[2a_1+(n-1)d\right] \\ {S}_2=\frac{10}{2}\left[2+18\right] \\ {S}_2=100 \end{gathered}$$

Step 4. Calculate the sum of series $S$.

$$\begin{gathered} \because S=S_1-S_2 \\ \Rightarrow S=110-100 \\ \Rightarrow S=10 \end{gathered}$$

Hence, the value of the given series is $10$.