Question referring to percentage increase

Question

Why if you evaluate the percentage increase between the first and last number of a sequence it doesn't equalize with the partial sum of the increases? Suppose you have the sequence $1,3,6$. The percentage increase of 1 to 6 is 500%. But if you do the partial sum of the increases, let's say 200% (from 1 to 3) and 100% (from 3 to 6), you get 300% and not 500%. Why is that? Furthermore, is there a way to make the partial sum of the increase equal the first to the nth number of a sequence? Thanks!

Answer 1

Let $A$ be the initial amount and assume there is a $20%$ percent increase from each term to the next term. Then the values you get are:

$$A, \;\;\;\; (1.2)A, \;\;\;\; (1.2)^2A, \;\;\;\; (1.2)^3A, \;\;\;\; \ldots$$

Notice that the 2nd term is $(1+0.2)A = A + (0.2)A,$ which is $20$% greater in magnitude than $A.$

However, the 3rd term is

$$(1+0.2)^2A \; = \; (1+0.2)(1+0.2)A \; = \; (1 + 0.2 + 0.2 + 0.04)A $$ $$= \; A + (0.2)A + (0.2)A + (0.04)A,$$

which is $20$% plus $20$% plus $4$% greater than $A.$

Also, the 4th term is

$$(1+0.2)^3A \; = \; (1+0.2)(1+0.2)(1 + 0.2)A$$ $$ = \; (1 + 0.2 + 0.2 + 0.2 + 0.04 + 0.04 + 0.04 + 0.008)A $$ $$= \; A + (0.2)A + (0.2)A + (0.2)A + (0.04)A + (0.04)A + (0.04)A + (0.008)A,$$

which is $20$% plus $20$% plus $20$% plus $4$% plus $4$% plus $4$% plus $0.8$% greater than $A.$

One way to explain the underlying reason why the 2nd term is not two $20$% increases, and the 3rd term is not three $20$% increases, etc. is due to the presence of the cross terms when expanding a binomial (i.e. $(x+y)^n$ is not just $x^n+y^{n}).$ Another way to explain this is that when computing the 3rd term, you increase the 2nd term by $20$%, so the 3rd term is the 2nd term plus $20$% of the 2nd term, which is a little more than the 2nd term plus $20$% of the 1st term; and when computing the 4th term, you increase the 3rd term by $20$%, so the 4th term is the 3rd term plus $20$% of the 3rd term, etc.