What is the formula for the #n#th term of the sequence #1, 3, 6, 10, 15,...# ?

2 Answers
Jan 25, 2017

Answer:

#a_n = 1/2n(n+1)#

Explanation:

These are triangular numbers:

#0color(white)(0000)0color(white)(000000)0color(white)(00000000)0color(white)(0000000000)0#
#color(white)(0000)0color(white)(0)0color(white)(0000)0color(white)(0)0color(white)(000000)0color(white)(0)0color(white)(00000000)0color(white)(0)0#
#color(white)(0000000000)0color(white)(0)0color(white)(0)0color(white)(0000)0color(white)(0)0color(white)(0)0color(white)(000000)0color(white)(0)0color(white)(0)0#
#color(white)(000000000000000000)0color(white)(0)0color(white)(0)0color(white)(0)0color(white)(0000)0color(white)(0)0color(white)(0)0color(white)(0)0#
#color(white)(0000000000000000000000000000)0color(white)(0)0color(white)(0)0color(white)(0)0color(white)(0)0#

Geometrically, you can see that such a triangle is one half of a parallelogram with base #n+1# and height #n#, e.g. #a_5:#

#color(white)(0000)0color(white)(0)color(blue)(0)color(white)(0)color(blue)(0)color(white)(0)color(blue)(0)color(white)(0)color(blue)(0)color(white)(0)color(blue)(0)#
#color(white)(000)0color(white)(0)0color(white)(0)color(blue)(0)color(white)(0)color(blue)(0)color(white)(0)color(blue)(0)color(white)(0)color(blue)(0)#
#color(white)(00)0color(white)(0)0color(white)(0)0color(white)(0)color(blue)(0)color(white)(0)color(blue)(0)color(white)(0)color(blue)(0)#
#color(white)(0)0color(white)(0)0color(white)(0)0color(white)(0)0color(white)(0)color(blue)(0)color(white)(0)color(blue)(0)#
#0color(white)(0)0color(white)(0)0color(white)(0)0color(white)(0)0color(white)(0)color(blue)(0)#

Such a parallelogram has a total count of #n(n+1)#, so the triangle is half of that.

In other words:

#a_n = sum_(k=1)^n k = 1/2n(n+1)#

Jan 25, 2017

Answer:

#(n^2+n)/2#

Explanation:

Starting with 1 you can begin to see a pattern develop....

#a_1=1#

#a_2=a_1+2=1+2=3#

#a_3=a_2+3=3+3=6#

#a_4=a_3+4=6+4=10#

#a_5=a_4+5=10+5=15#

#vdots#

#a_n=a_(n-1)+n#

Since every term is merely the sum of the previous term plus the next counting number we have.

#a_n=sum_(k=1)^nk=(n(n+1))/2#

To test this we use the first formula for #a_(n+1)#

#a_(n+1)=a_n+(n+1)#

#=> a_(n+1)=(n(n+1))/2+(n+1)=(n^2+n)/2+(2n+2)/2=(n^2+3n+2)/2#

by the hypothesis

#a_(n+1)=((n+1)(n+2))/2=(n^2+3n+2)/2#

They are equal...

So the hypothesis is proven

This is an example of "Weak" Induction