Animated Solution for Mathematics - Sequence and Series: If $1+(1-2^2\cdot 1)+(1-4^2\cdot 3)+(1-6^2\cdot 5)+......+(1-20^2\cdot 19)=\alpha -220\beta$, then an ordered pair $(\alpha ,\beta )$ is equal to :

Imagine you are standing before a complex series, a wall of numbers that seems chaotic at first glance: $1+(1-2^2\cdot 1)+(1-4^2\cdot 3)+(1-6^2\cdot 5)+\cdots +(1-20^2\cdot 19)$. Many students see this and panic, but an elite JEE aspirant sees a story waiting to be told.

The secret to solving any series problem is to find its DNA—the general term. Notice the isolated $1$ at the start; it is a constant, a sentinel guarding the rest of the series.

Now, look at the brackets. Each one follows a beautiful, predictable rhythm: $1$ minus an even number squared, multiplied by an odd number that is exactly one less than that even number. This is the key to unlocking the entire problem.

Let us define the general term $T_r$ for the bracketed part. The even numbers being squared are $2,4,6,\dots ,20$. We can represent any such even number as $2r$.

The multipliers are $1,3,5,\dots ,19$, which are simply $(2r-1)$. Thus, the general term for the brackets is:

When we expand this, we get $T_r=1-4r^2(2r-1)$, which simplifies to:

This is the heart of the problem. By identifying this, we have transformed a terrifying string of numbers into a clean, manageable algebraic expression.

Now, we must sum these terms. The series $S$ is $1+{\sum }_{r=1}^{10}{T}_r$. Substituting our expanded $T_r$, we get:

We can distribute the summation operator to make it even simpler:

This is where your toolkit of standard summation formulas becomes your greatest weapon. You know that:

For $n=10$, these calculations are straightforward. The sum of squares is:

The sum of cubes is:

With these values in hand, we substitute them back into our expression for $S$:

This gives us $S=11-24200+1540$. Adding the positive terms, we get $1551-24200$, which equals $-22649$.

We are told that $S=\alpha -220\beta$. So, $-22649=\alpha -220\beta$. By checking the options, we test $\alpha =11$, which leads to:

Dividing $22660$ by $220$ gives us $\beta =103$. The ordered pair $(\alpha ,\beta )$ is $(11,103)$.