Animated Solution for Mathematics - Sequence and Series: The greatest integer less than or equal to the sum of first 100 terms of the sequence $\frac{1}{3},\frac{5}{9},\frac{19}{27},\frac{65}{81},\dots$ is equal to

My dear student, welcome to a journey of mathematical discovery. Today, we are going to unravel a sequence that might look intimidating at first, but hides a beautiful, elegant structure.

We are given the sequence:

When you first see these fractions, it is natural to feel a bit lost. The numerators—$1,5,19,65$—do not immediately scream a simple arithmetic or geometric progression.

But in mathematics, when the surface is chaotic, the secret often lies in the foundation. Look at the denominators: $3,9,27,81$. These are clearly powers of $3$: $3^1,3^2,3^3,3^4$. This is our first breakthrough!

Now, let us look at the relationship between the numerator and the denominator. For the first term, $T_1=\frac{1}{3}$, we can write it as $\frac{3-2}{3}$, which simplifies to $1-\frac{2}{3}$.

For the second term, $T_2=\frac{5}{9}$, we can write it as $\frac{9-4}{9}$, which is $1-\frac{4}{9}$, or $1-(\frac{2}{3}{)}^2$.

For the third term, $T_3=\frac{19}{27}$, we see $\frac{27-8}{27}$, which is $1-\frac{8}{27}$, or $1-(\frac{2}{3}{)}^3$.

Do you see the elegance emerging? The general term $T_n$ is simply: This is the key that unlocks the entire problem.

We need to find the sum of the first $100$ terms, $S_{100}={\sum }_{n=1}^{100}{T}_n$. Substituting our general term, we get:

Using the linearity of summation, we can split this into two distinct parts:

The first part is trivial: adding $1$ to itself $100$ times gives us $100$. The second part is a classic Geometric Progression (GP) where the first term $a=\frac{2}{3}$ and the common ratio $r=\frac{2}{3}$.

Using the GP sum formula $S=\frac{a(1-r^n)}{1-r}$, we calculate the sum of the second part:

Since $1-\frac{2}{3}=\frac{1}{3}$, the expression simplifies beautifully to $2(1-(\frac{2}{3}{)}^{100})$. Now, bringing it all together:

We are almost there! We need the greatest integer less than or equal to $S_{100}$. We have established that $S_{100}=98+2(\frac{2}{3}{)}^{100}$.

Since $(\frac{2}{3}{)}^{100}$ is a very small positive number, $2(\frac{2}{3}{)}^{100}$ is also a small positive number strictly between $0$ and $1$. Thus, $S_{100}$ is $98$ plus a tiny fraction.

By the definition of the greatest integer function, $\lfloor 98+ϵ\rfloor =98$. And there you have it—the complexity dissolves into a simple, satisfying integer. The final answer is 98.