A sequence of numbers that are derived based on some mathematical relationship between terms is called a mathematical series.

Series can be of different types, two of which we study here.

So the series of n terms here is: $a, a+d, a+2d, ..., a+(n-1)*d$

The sum of the first n terms is: $a+(a+d)+(a+2d)+...+(a+(n-1)*d)=n*a+d*(1+2+3+...+(n-1))=n*a+d*(n-1)*n/2=(n/2)*[2a+(n-1)*d]$

Here, we use the fact that the sum of the first n consecutive natural numbers = $n*(n+1)/2$.

This can be proved easily using either mathematical induction or the method of finite differences both of which are currently out of scope of this tutorial.

e.g. the series $1, 3, 5, 7, 9, ..., 31. 31=1+15*2$, so $1+3+...+31=16*(2+15*2)/2=256$

Other useful math formulas:

The sum of the squares of the first n consecutive natural numbers = $n*(n+1)*(2n+1)/6$.

The sum of the cubes of the first n consecutive natural numbers=the square of the sum of the first n consecutive natural numbers=$(n*(n+1)/2)^2$

So the series of n terms here is: $a, a*r, a*r*r, a*r*r*r, .... a*r^(n-1)$.

The sum of the first n terms of this series is given by the formula $S=a*(r^n-1)/(r-1)$ if $r>1$ or $a*(1-r^2)/(1-r)$ if $r<1$.

e.g. $1,2,4,8,16,...1024$. So $a=1, r=2, 1024=2^10$, so $n=11$. $S=1*(2^11-1)/(2-1)=2047$

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