The sum of the series 1 + 2 × 3 + 3 × 5 + 4 × 7 + .... upto 11th term is :- (A) 915 (B) 946 (C) 945

Q: Wait, why did we rewrite the first term as 1 x 1? Is that even allowed?

Absolutely! In mathematics, especially when dealing with series, we look for a 'general form' that governs every term. By writing $1$ as $1 \times 1$, we reveal the underlying pattern of the sequence: every term is a product of two numbers. It makes the logic consistent for all $r$.

Q: How do I know if the second factor is an Arithmetic Progression?

Great question! Just look at the difference between consecutive terms: $3-1=2$, $5-3=2$, $7-5=2$. Since the difference is constant, it is by definition an Arithmetic Progression. Once you identify the first term $a=1$ and common difference $d=2$, the formula $a + (r-1)d$ will always give you the $r^{th}$ term.

Q: Can I just add the 11 terms manually instead of using the summation formulas?

You could, but in the JEE, time is your most precious resource. Manual addition is prone to calculation errors and is inefficient. Mastering the summation formulas $\sum r^2$ and $\sum r$ is a hallmark of a prepared student—it turns a 5-minute tedious task into a 30-second elegant calculation.

Q: Why is the formula for $\sum r^2$ divided by 6? Where does that come from?

That is a fantastic curiosity! The formula $\frac{n(n+1)(2n+1)}{6}$ is derived using the method of differences or mathematical induction. While you don't need to derive it during the exam, knowing that it represents the sum of squares helps you visualize the growth of the series.

Q: I got 946, but the options say 946 is the second one. Did I miss something?

Don't panic! If your calculation matches the result, you have succeeded. Always double-check your arithmetic in the final subtraction step ($1012 - 66$). If you arrived at 946, you have mastered the logic of this series!

Question

The sum of the series $1 + 2 	imes 3 + 3 	imes 5 + 4 	imes 7 + \dots$ upto 11th term is :-

Options:

(A) 915

(B) 946

(C) 945

(D) 916

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a wall of numbers: $1 + 2 	imes 3 + 3 	imes 5 + 4 	imes 7 + \dots$. At first glance, it looks like a chaotic jumble of products.

In the world of JEE Advanced, chaos is just order waiting to be discovered. To solve this, we don't just calculate; we observe the heartbeat of the series.

### Decoding the DNA of the Series

Every series has a 'general term', a mathematical DNA that dictates how every single piece of the puzzle behaves. Let's look at the first factors: $1, 2, 3, 4, \dots$. This is the simplest sequence in existence—the natural numbers, where the $r^{th}$ term is simply $r$.

Now, look at the second factors: $1, 3, 5, 7, \dots$. This is an Arithmetic Progression (A.P.) where each step increases by $2$.

Using the formula for the $r^{th}$ term of an A.P., $T_r = a + (r-1)d$, we plug in $a=1$ and $d=2$ to get $1 + (r-1)2$, which simplifies to $2r - 1$. By multiplying these two, we find the general term of our series:

$$T_r = r(2r - 1) = 2r^2 - r$$

This is the key that unlocks the entire problem. We have transformed a confusing list of numbers into a clean, algebraic expression.

### The Power of Summation

Now that we have $T_r = 2r^2 - r$, we need to find the sum of the first $11$ terms. We express this as:

$$S_{11} = \sum_{r=1}^{11} (2r^2 - r)$$

Linearity is our best friend here. We can split this into two manageable parts:

$$S_{11} = 2 \sum_{r=1}^{11} r^2 - \sum_{r=1}^{11} r$$

This is where the magic of standard formulas comes into play. We know that the sum of the first $n$ squares is given by:

$$\sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}$$

Additionally, the sum of the first $n$ natural numbers is:

$$\sum_{r=1}^{n} r = \frac{n(n+1)}{2}$$

### The Final Calculation

Let's substitute $n=11$ into our expressions. For the sum of squares:

$$2 	imes \frac{11(11+1)(2 	imes 11 + 1)}{6} = 2 	imes \frac{11 	imes 12 	imes 23}{6}$$

The $6$ in the denominator cancels perfectly with the $12$ in the numerator, leaving us with $2 	imes 11 	imes 2 	imes 23$. That gives us $44 	imes 23 = 1012$.

Next, we calculate the sum of the natural numbers:

$$\sum_{r=1}^{11} r = \frac{11(12)}{2} = 11 	imes 6 = 66$$

Finally, we bring it all together. The total sum is the difference between these two values:

$$S_{11} = 1012 - 66 = 946$$

### Conclusion: The Elegance of Logic

We took a series that seemed daunting and broke it down into its fundamental components. We identified the patterns, applied the correct tools, and arrived at the answer with precision.

This is the essence of JEE preparation—not memorizing solutions, but learning to see the underlying structure of the universe. The final result is **946**.

Animated Solution for Mathematics - Sequence and Series: The sum of the series $1 + 2 \times 3 + 3 \times 5 + 4 \times 7 + \dots$ upto 11th term is :-

Select Answer:

Visualized Solution

Series Observation

First Factor Pattern

Second Factor Pattern

General Term $T_{r}$

Summation Setup $S_{11}$

Standard Summation Formulas

Substituting in $\sum r^{2}$

Evaluating $2 \sum r^{2}$

Evaluating $\sum r$

Final Subtraction

The Sigma Insight: Sum of Special Series

Analyzing the Setup

Decoding the DNA of the Series

The Power of Summation

The Final Calculation

Conclusion

Similar Questions

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Animated Solution for Mathematics - Sequence and Series: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum of the series 1+2×3+3×5+4×7+… upto 11th term is :-

Select Answer:

Visualized Solution

Series Observation

First Factor Pattern

Second Factor Pattern

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

Summation Setup .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S11​

Standard Summation Formulas

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Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2∑r2

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∑r