Animated Solution for Mathematics - Sequence and Series: ${\sum }_{r=1}^{20}(r^2+1)(r!)$ is equal to:

Mastering the Factorial Summation

Welcome, future engineer. Today, we are not just solving a math problem; we are uncovering a hidden rhythm within a series.

When you look at the expression ${\sum }_{r=1}^{20}(r^2+1)(r!)$, your first instinct might be to panic. Factorials grow explosively, and quadratic terms add a layer of complexity that seems to defy standard summation formulas.

But take a deep breath. In the world of JEE Advanced, whenever you see a factorial multiplied by a polynomial, you are almost certainly looking at a 'Telescoping Series' in disguise.

The Anatomy of the General Term

Let us isolate the general term, $T_r=(r^2+1)r!$. Our mission is to transform this into a difference of two consecutive terms, $V_{r+1}-V_r$.

Why? Because if we can do that, the summation becomes a beautiful, cascading collapse. Imagine a line of dominoes:

Notice how the positive $V_2$ cancels the negative $V_2$, the $V_3$ cancels the $V_3$, and so on. We are left only with the final term $V_{21}$ and the initial term $V_1$. This is the power of the Method of Differences.

The Algebraic Surgery

Now, how do we force $(r^2+1)$ to cooperate? We need to create an $(r+1)$ factor to turn $r!$ into $(r+1)!$.

Let us perform some algebraic surgery. We take $(r^2+1)$ and add and subtract $r$:

Grouping these, we get $(r^2+r)-(r-1)$. Factoring the first part gives us $r(r+1)-(r-1)$.

Now, watch what happens when we multiply this by $r!$:

Distributing the $r!$, we obtain $T_r=r(r+1)r!-(r-1)r!$. Using the fundamental property of factorials, where $(n+1)n!=(n+1)!$, the first term simplifies beautifully:

The Telescoping Magic

We have arrived at the breakthrough. Let us define our function $V_r=(r-1)r!$.

If we calculate $V_{r+1}$, we replace $r$ with $(r+1)$, giving us $((r+1)-1)(r+1)!$, which is simply $r(r+1)!$. Look at that! Our general term $T_r$ is exactly $V_{r+1}-V_r$.

The series is now ready to collapse. The sum $S={\sum }_{r=1}^{20}(V_{r+1}-V_r)$ expands to:

As we predicted, the intermediate terms vanish, leaving us with $S=V_{21}-V_1$.

The Final Polish

Calculating $V_1$ is trivial: $(1-1)1!=0$. Calculating $V_{21}$ is straightforward: $(21-1)21!=20\times 21!$.

Thus, our sum is $20\times 21!$. But wait, look at the options. We need to match our result to the provided choices.

We rewrite $20$ as $(22-2)$. Then:

Since $22\times 21!=22!$, our final answer is:

You have successfully navigated the complexity, manipulated the algebra, and emerged victorious. This is the essence of JEE Advanced physics and math: seeing the structure beneath the chaos. Keep this logic in your toolkit, and no series will ever intimidate you again.