Let Sn = 1 · (n - 1) + 2 · (n - 2) + 3 · (n - 3) + ... + (n - 1) · 1, n ≥ 4. The sum Σ(n=4 to ∞) (2S

Q: Wait, why does the summation for $S_n$ go up to $n-1$?

Look at the pattern: $1 \cdot (n-1) + 2 \cdot (n-2) + \dots + (n-1) \cdot 1$. The first factor $r$ goes from $1$ to $n-1$. When $r=1$, the second factor is $n-1$. When $r=n-1$, the second factor is $n-(n-1) = 1$. Thus, the summation index $r$ correctly ranges from $1$ to $n-1$.

Q: How do I know when to expand the factorial in the denominator?

The goal is to create terms that cancel out. Whenever you see a polynomial in $n$ in the numerator and a factorial in the denominator, try to express the factorial in terms of the factors present in the numerator. Here, $n! = n(n-1)(n-2)!$ allowed us to cancel $n(n-1)$ and simplify the expression significantly.

Q: Is there a trick to remember the $e$ series?

Yes! The Maclaurin series for $e^x$ is $\sum_{k=0}^{\infty} \frac{x^k}{k!}$. For $x=1$, this is $e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots$. Whenever you see a sum of reciprocals of factorials, always check if it matches this series. If the first few terms are missing, just subtract them from $e$.

Question

Let $S_n = 1 \cdot (n - 1) + 2 \cdot (n - 2) + 3 \cdot (n - 3) + \dots + (n - 1) \cdot 1, n \ge 4$. The sum $\sum_{n=4}^{\infty} \left( \frac{2 S_n}{n!} - \frac{1}{(n-2)!} \right)$ is equal to :

Options:

(A) $\frac{e-1}{3}$

(B) $\frac{e-2}{6}$

(C) $\frac{e}{3}$

(D) $\frac{e}{6}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the JEE journey. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of factorials and summations.

As we peel back the layers, you will see that it is actually a beautifully orchestrated dance of algebra and series. Let us begin by looking at the structure of $S_n = 1 \cdot (n - 1) + 2 \cdot (n - 2) + 3 \cdot (n - 3) + \dots + (n - 1) \cdot 1$.

The first thing to notice is the symmetry. We have a product of two terms where the first increases and the second decreases. We can define the general term as $T_r = r(n - r)$, where $r$ ranges from $1$ to $n-1$.

By writing $S_n = \sum_{r=1}^{n-1} (nr - r^2)$, we have successfully translated the verbal description into the language of mathematics. Now, we use the linearity of summation to split this into two parts:

$$S_n = n \sum_{r=1}^{n-1} r - \sum_{r=1}^{n-1} r^2$$

This is where many students stumble—the limits. Remember, the sum goes up to $n-1$, not $n$. Applying the standard formulas, we get:

$$S_n = n \frac{(n-1)n}{2} - \frac{(n-1)n(2n-1)}{6}$$

With a bit of algebraic finesse, factoring out $\frac{n(n-1)}{2}$, we find that:

$$S_n = \frac{n(n-1)(n+1)}{6}$$

### The Factorial Dance

Now that we have tamed $S_n$, let us look at the infinite series: $\sum_{n=4}^{\infty} \left( \frac{2 S_n}{n!} - \frac{1}{(n-2)!} \right)$. Let us substitute our simplified $S_n$ into the general term $U_n$:

$$U_n = \frac{2 \left( \frac{n(n-1)(n+1)}{6} \right)}{n!} - \frac{1}{(n-2)!}$$

The $2$ and $6$ cancel to leave a $3$ in the denominator. Now, the magic happens. We know that $n! = n(n-1)(n-2)!$.

By substituting this into the first term, the $n(n-1)$ terms in the numerator and denominator vanish! We are left with:

$$U_n = \frac{n+1}{3(n-2)!} - \frac{1}{(n-2)!}$$

Combining these over a common denominator, we get:

$$U_n = \frac{n+1-3}{3(n-2)!} = \frac{n-2}{3(n-2)!}$$

### The Grand Finale

We are almost there. Notice that $\frac{n-2}{(n-2)!}$ is simply $\frac{1}{(n-3)!}$. So, our general term is $U_n = \frac{1}{3(n-3)!}$.

The infinite sum is now:

$$S = \frac{1}{3} \sum_{n=4}^{\infty} \frac{1}{(n-3)!}$$

Let $k = n-3$. As $n$ goes from $4$ to $\infty$, $k$ goes from $1$ to $\infty$. Thus:

$$S = \frac{1}{3} \sum_{k=1}^{\infty} \frac{1}{k!} = \frac{1}{3} \left( \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots \right)$$

We know that $e = 1 + \frac{1}{1!} + \frac{1}{2!} + \dots$, so the sum in the parentheses is $e - 1$. Therefore, the final answer is:

$$\mathbf{\frac{e-1}{3}}$$

It is elegant, it is clean, and it is a testament to the power of simplification. Keep practicing, and soon, you will see these patterns before you even pick up your pen.

Animated Solution for Mathematics - Sequence and Series: Let $S_{n} = 1 \cdot (n - 1) + 2 \cdot (n - 2) + 3 \cdot (n - 3) + \dots + (n - 1) \cdot 1, n \geq 4$ . The sum $\sum_{n = 4}^{\infty} (\frac{2 S _{n}}{n !} - \frac{1}{( n - 2 )!})$ is equal to :

Select Answer:

Visualized Solution

General Term of $S_{n}$

Expanding the Summation

Applying Summation Formulas

Substituting Limits into Formulas

Algebraic Simplification of $S_{n}$

Final Form of $S_{n}$

Setting up the Infinite Sum

Substituting $S_{n}$ into $U_{n}$

Simplifying the Factorial Expression

Combining the Fractions

Further Simplifying the Factorial

Setting up the Infinite Series

Expanding the Series

Relating to the Exponential Series $e$

Final Conclusion

The Sigma Insight: Sum of Special Series

Analyzing the Setup

The Factorial Dance

The Grand Finale

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Select Answer:

Visualized Solution

General Term of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Sn​

Expanding the Summation

Applying Summation Formulas

Substituting Limits into Formulas

Algebraic Simplification of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Sn​

Final Form of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Sn​