Animated Solution for Mathematics - Sequence and Series: Comprehension Passage

Imagine you are standing in front of a sequence generator. Usually, an A.P. is fixed: $a$ is $2$, $d$ is $3$, and you just sum them up. But here, the generator is dynamic.

The first term $a$ is $r$, and the common difference $d$ is $2r-1$. We are asked to find $V_r$, the sum of the first $r$ terms.

We pull our trusty toolkit out: the sum of an A.P. is given by $S_n=\frac{n}{2}[2a+(n-1)d]$. Substituting our variables, we get:

Take a deep breath—don't rush the algebra. Expanding the inner terms, $(r-1)(2r-1)$ becomes $2r^2-3r+1$. Adding the $2r$ from the first part, we get $2r^2-r+1$.

Finally, multiplying by $\frac{r}{2}$, we arrive at the elegant cubic form:

This is the heartbeat of our problem.

Now, we need the sum of $V_r$ from $r=1$ to $n$. This is where we use the linearity of summation. We are looking for:

We split this into three manageable pieces: $\sum r^3-\frac{1}{2}\sum r^2+\frac{1}{2}\sum r$.

Recall your standard formulas: $\sum r^3=\frac{n^2(n+1{)}^2}{4}$, $\sum r^2=\frac{n(n+1)(2n+1)}{6}$, and $\sum r=\frac{n(n+1)}{2}$. Substituting these in, we get a seemingly terrifying expression.

Look closely—there is a common factor of $\frac{n(n+1)}{12}$ waiting to be pulled out. Once you factor that out, the expression inside the bracket simplifies beautifully to $3n^2+n+2$.

The result is:

See? The chaos resolves into order.

Next, we define $T_r=V_{r+1}-V_r-2$. This is essentially the difference between consecutive terms of our $V$ sequence.

By substituting our cubic polynomial for $V_r$ and $V_{r+1}$, the $r^3$ terms cancel out, leaving us with a quadratic:

Is it prime? Is it composite? Let's factorize it. It splits into $(3r-1)(r+1)$.

For any $r\ge 1$, the factor $(3r-1)$ is at least $2$, and $(r+1)$ is at least $2$. A number that is the product of two integers greater than $1$ is, by definition, a composite number. We have solved the mystery!

Finally, we define $Q_r=T_{r+1}-T_r$. Since $T_r$ is a quadratic, its first difference must be linear.

Let's calculate:

Expanding this, the $3r^2$ terms vanish, and we are left with:

This is a linear function of $r$. Any sequence whose general term is linear in $r$ is an Arithmetic Progression.

The common difference is simply the coefficient of $r$, which is $6$. Thus, $Q_r$ is an A.P. with a common difference of $6$.