# Prove by induction the formula for the sum of the first n terms of an arithmetic series.

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The simplest and most common form of mathematical induction infers that a statement involving a natural number

nholds for all values ofn. The proof consists of two steps:

- The
basis(base case): prove that the statement holds for the first natural numbern. Usually,n= 0 orn= 1.- The
inductive step: prove that, if the statement holds for some natural numbern, then the statement holds forn+ 1.The hypothesis in the inductive step that the statement holds for some

nis called theinduction hypothesis(orinductive hypothesis). To perform the inductive step, one assumes the induction hypothesis and then uses this assumption to prove the statement forn+ 1.Whether

n= 0 orn= 1 depends on the definition of the natural numbers. If 0 is considered a natural number, as is common in the fields of combinatorics and mathematical logic, the base case is given byn= 0. If, on the other hand, 1 is taken as the first natural number, then the base case is given byn= 1.

The set of natural numbers as used by mathematicians today includes 0.

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The formula for the sum of the first n terms of an arithmetic series is `S_n = (n/2)*(2a + (n-1)*d)`

To prove this by mathematical induction, first determine `S_0` , this is equal to 0. Which follows from the fact the the sum of zero terms is 0.

Let `S_n = (n/2)*(2a + (n-1)*d)` be true , it has to be shown that `S_(n+1) = S_n + T_(n+1)`

S_(n+1) = `((n+1)/2)*(2a + n*d)`

= `(n/2)*(2a + (n -1)*d + d) + (1/2)*(2a + n*d)`

= `(n/2)*(2a + (n -1)*d) + (n/2)*d + (1/2)*(2a + n*d)`

= `S_n + a + (n/2)*d + (n/2)*d`

= `S_n + a + n*d`

= `S_n + T_(n+1)`

This shows that the formula holds for n = 0 and if it holds for n it also holds for n + 1.

**The formula for the sum of the first n terms of an arithmetic series is `S_n = (n/2)*(2a + (n-1)*d)` **

The sum of a arithmetic series is given by;

`S_n = n/2(2a+(n-1)d) `

where;

a = initial term

d = common difference

n = number of terms

So the arithmetic series will have a look like the follows.

`a,(a+d),(a+2d),(a+3d)............,(a+(n-1)d)`

Now let us prove this using mathematical induction.

What we need to do is to prove the result for smallest of n which is n = 1 and assume for n = p (p>1) result is true and prove the result for n = p+1

When n = 1 the number of terms will be 1 or the only term in the progression is initial term 'a'.

`S_n = n/2(2a+(n-1)d) `

`S_1 = 1/2(2a+(1-1)d)`

`S_1 = a`

For n = 1 result is true.

Let us assume for n = p the result is true.

`S_p = p/2(2a+(p-1)d)`

The nth term of a arithmetic series is given by;

`T_n = a+(n-1)d`

So we can say the last term of the arithmetic series with p terms as;

`T_p = a+(p-1)d`

Similarly we can say;

`T_(p+1) = a+((p+1)-1)d = a+pd`

When we consider the sum of (p+1) terms;

`S_(p+1) = S_p+T_(p+1)`

`S_(p+1) = p/2(2a+(p-1)d)+(a+pd)`

`S_(p+1) = ap+p(p-1)d/2+a+pd`

`S_(p+1) = a(p+1)+pd((p-1)/2+1)`

`S_(p+1) = a(p+1)+(pd)/2(p+1)`

`S_(p+1) = (p+1)/2(2a+pd)`

`S_(p+1)= (p+1)/2(2a+((p+1)-1)d)`

So for n = p+1 the result is true.

*So from mathematical induction for every `ninZ^+` the result is true.*

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