Question 637274
The terms are in a geometric sequence, or geometric progression.
Each term is the previous one multiplied times {{{4}}} .
You may be studying geometric sequences, also called geometric progressions.
The symbols and names used vary a little from place to place, from book to book, and from teacher to teacher, so the symbols, and formulas may look different, but the idea is the same. I will use a popular choice for terms, symbols, and formulas. Keep in mind that your book/teacher may use slightly different ones.
 
THE GENERAL THEORY (what the book may say):
In a geometric sequence, or geometric progression, the term number {{{n}}} and the number {{{n-1}}} (the previous one) are related by the recurrence formulas
{{{a[n]=a[n-1]*r}}} or {{{r=a[n]/(a[n-1])}}} ,
Where {{{a[n]}}} is the tern number {{{n}}} ,
{{{a[n-1]}}} is the previous term, and {{{r}}} is the common ratio.
The term number {{{n}}} can be calculated from the first term and the common ratio, using the explicit formula
{{{a[n]=a[1]*r^(n-1)}}}
The sum of the first {{{n}}} terms, {{{S[n]=a[1]+a[2]}}}+ ... +{{{a[n-1] + a[n]}}}
can be calculated as
{{{S[n]=a[1](r^n-1)/(r-1)}}}
The formula can be deduced as follows:
{{{S[n]=a[1]+r*a[1]+r^2*a[1]}}}+ ... +{{{r^(n-2)*a[1]+r^(n-1)*a[1]}}} , so
{{{r*S[n]=r*a[1]+r^2*a[1]}}}+ ... +{{{r^n*a[1]+r^(n-1)*a[1]+r^n*a[1]}}}
Subtracting one equation from the other, a lot of terms cancel, and you have
{{{r*S[n]-S[n]=r^n*a[1]-a[1]}}} --> {{{(r-1)*S[n]=(r^n-1)*a[1]}}}
Then, dividing both sides by {{{r-1}}}, we get
{{{(r-1)*S[n]=(-1)r^n*a[1]}}} --> {{{S[n]=a[1](r^n-1)/(r-1)}}}
Someone discovered that formula. I just remember the trick and re-discover the formula every time, because I do not need to use it often, and do not remember it.
 
THE SOLUTION TO THE PROBLEM:
{{{a[1]=1}}}, {{{r=4}}}, {{{n=10}}}
Applying the formula {{{S[n]=a[1](r^n-1)/(r-1)}}} , we get
{{{S[n]=1*(4^10-1)/(4-1)=(4^10-1)/3=(1048576-1)/3=1048575/3=highlight(349525)}}}