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Hint: In this question it is given that we have to find the value of the given series,

0-1+2-3+4-5+6-7+...........+16-17+18-19+20.

So to find the solution we need to know that if any series: a+(a+d)+(a+2d)+.......

is in A.P(Arithmetic Progression) then the summation of first n terms is,

$$S=\dfrac{n}{2} \left( 2a+\left( n-1\right) d\right) $$.......(1)

Where a and d are the first term and common difference respectively,

So in order to get the solution we have to rearrange the given series.

So the given series,

0-1+2-3+4-5+6-7+...........+16-17+18-19+20

Now we are going to rearrange the positions by taking all the negative terms in one side,

So we can write,

0-1+2-3+4-5+6-7+...........+16-17+18-19+20

=0+2+4+6+8+10+12+14+16+18+20-1-3-5-7-9-11-13-15-17-19

=(2+4+6+8+10+12+14+16+18+20)-(1+3+5+7+9+11+13+15+17+19)

=$$S_{1}-S_{2}$$(say)

Where,

$$S_{1}=2+4+6+8+10+\ldots +20$$

$$S_{2}=1+3+5+7+9+11+\ldots +19$$

Now for $$S_{1}$$ the first term is a=2 and common difference, d=2-0=2 and total number of elements are n=10,

Therefore by formula (1) we can write,

$$S_{1}=\dfrac{n}{2} \left( 2a+\left( n-1\right) d\right) $$

$$=\dfrac{10}{2} \left( 2\times 2+\left( 10-1\right) \times 2\right) $$

$$=5\times \left( 4+9\times 2\right) $$

$$=5\times 22$$

$$=110$$

Now similarly for $$S_{2}$$ the first term is a=1 and common difference, d=3-1=2 and total number of elements are n=10,

Therefore by formula (1) we can write,

$$S_{2}=\dfrac{n}{2} \left( 2a+\left( n-1\right) d\right) $$

$$=\dfrac{10}{2} \left( 2\times 1+\left( 10-1\right) \times 2\right) $$

$$=5\times \left( 2+9\times 2\right) $$

$$=5\times \left( 2+18\right) $$

$$=5\times 20$$

$$=100$$

So we can write,

$$S_{1}-S_{2}=110-100=10$$

Therefore, the value of 0-1+2-3+4-5+6-7+...........+16-17+18-19+20 is 10.

Note: While solving any series related problem you need to first check whether the given series follows any order or not (A.P, G.P or H.P), if not then you have to make it by doing rearrangement like we did while solving the above problem. Where after rearranging we get two series which are in G.P.

0-1+2-3+4-5+6-7+...........+16-17+18-19+20.

So to find the solution we need to know that if any series: a+(a+d)+(a+2d)+.......

is in A.P(Arithmetic Progression) then the summation of first n terms is,

$$S=\dfrac{n}{2} \left( 2a+\left( n-1\right) d\right) $$.......(1)

Where a and d are the first term and common difference respectively,

So in order to get the solution we have to rearrange the given series.

__Complete step-by-step solution:__So the given series,

0-1+2-3+4-5+6-7+...........+16-17+18-19+20

Now we are going to rearrange the positions by taking all the negative terms in one side,

So we can write,

0-1+2-3+4-5+6-7+...........+16-17+18-19+20

=0+2+4+6+8+10+12+14+16+18+20-1-3-5-7-9-11-13-15-17-19

=(2+4+6+8+10+12+14+16+18+20)-(1+3+5+7+9+11+13+15+17+19)

=$$S_{1}-S_{2}$$(say)

Where,

$$S_{1}=2+4+6+8+10+\ldots +20$$

$$S_{2}=1+3+5+7+9+11+\ldots +19$$

Now for $$S_{1}$$ the first term is a=2 and common difference, d=2-0=2 and total number of elements are n=10,

Therefore by formula (1) we can write,

$$S_{1}=\dfrac{n}{2} \left( 2a+\left( n-1\right) d\right) $$

$$=\dfrac{10}{2} \left( 2\times 2+\left( 10-1\right) \times 2\right) $$

$$=5\times \left( 4+9\times 2\right) $$

$$=5\times 22$$

$$=110$$

Now similarly for $$S_{2}$$ the first term is a=1 and common difference, d=3-1=2 and total number of elements are n=10,

Therefore by formula (1) we can write,

$$S_{2}=\dfrac{n}{2} \left( 2a+\left( n-1\right) d\right) $$

$$=\dfrac{10}{2} \left( 2\times 1+\left( 10-1\right) \times 2\right) $$

$$=5\times \left( 2+9\times 2\right) $$

$$=5\times \left( 2+18\right) $$

$$=5\times 20$$

$$=100$$

So we can write,

$$S_{1}-S_{2}=110-100=10$$

Therefore, the value of 0-1+2-3+4-5+6-7+...........+16-17+18-19+20 is 10.

Note: While solving any series related problem you need to first check whether the given series follows any order or not (A.P, G.P or H.P), if not then you have to make it by doing rearrangement like we did while solving the above problem. Where after rearranging we get two series which are in G.P.

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