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

The simplest and most common form of mathematical induction infers that a statement involving a natural number n holds for all values of n. The proof consists of two steps:

1. The basis (base case): prove that the statement holds for the first natural number n. Usually, ...

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The simplest and most common form of mathematical induction infers that a statement involving a natural number n holds for all values of n. The proof consists of two steps:

1. The basis (base case): prove that the statement holds for the first natural number n. Usually, n = 0 or n = 1.
2. The inductive step: prove that, if the statement holds for some natural number n, then the statement holds for n + 1.

The hypothesis in the inductive step that the statement holds for some n is called the induction hypothesis (or inductive hypothesis). To perform the inductive step, one assumes the induction hypothesis and then uses this assumption to prove the statement for n + 1.

Whether n = 0 or n = 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 by n = 0. If, on the other hand, 1 is taken as the first natural number, then the base case is given by n = 1.

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

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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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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)`

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