JEE Main 2021 (March)

LEVELJEE Main

Animated Solution for Mathematics - Sequence and Series: If $\sum_{r = 1}^{10} r! (r^{3} + 6 r^{2} + 2 r + 5) = α (11!)$ , then the value of $α$ is equal to

Q: Why do we use $(r+1)(r+2)(r+3)$? Why not just $(r+1)(r+2)$?

Great question! The degree of your polynomial is 3 ($r^3$). To 'absorb' a cubic term into a factorial, you need three consecutive factors to step up the factorial from $r!$ to $(r+3)!$. If the polynomial were quadratic, you would only need two. Always match the number of consecutive factors to the degree of the polynomial.

Enter Numerical Value:

Progressive Hints

Hint 1

Look at the structure of the term

r! (r^{3} + 6 r^{2} + 2 r + 5)

. When you see a factorial multiplied by a polynomial, your goal is to transform the polynomial into a form that allows you to create higher factorials. Think about how

(r + 1) (r + 2) (r + 3) r!

becomes

(r + 3)!

. Can you decompose the cubic polynomial to make this happen?

Hint 2

You need to express

r^{3} + 6 r^{2} + 2 r + 5

as a combination of

(r + 1) (r + 2) (r + 3)

(r + 1)

, and a constant. Start by expanding

(r + 1) (r + 2) (r + 3)

and compare it to your polynomial. The difference will be a linear expression in

r

. How can you rewrite that linear expression to involve

(r + 1)

Hint 3

Once you have the decomposition

P (r) = (r + 1) (r + 2) (r + 3) - 9 (r + 1) + 8

, substitute this back into the summation. You will get

\sum [(r + 3)! - 9 (r + 1)! + 8 (r!)]

. Now, use the linearity of the summation to write this as three separate sums. The key is to write out the first few and last few terms of each sum to see how the intermediate terms cancel out.

Visualized Solution

$\sum_{r = 1}^{10} r! (r^{3} + 6 r^{2} + 2 r + 5) = α (11!)$

Given equation: $\sum_{r = 1}^{10} r! (r^{3} + 6 r^{2} + 2 r + 5) = α (11!)$
Objective: Find the value of the constant $α$ .
Key Strategy: Decompose the polynomial $r^{3} + 6 r^{2} + 2 r + 5$ into terms that can be combined with $r!$ to form higher factorials.

Expressing $P (r)$ in terms of $(r + 1), (r + 2), (r + 3)$

We want to express $P (r) = r^{3} + 6 r^{2} + 2 r + 5$ in terms of $(r + 1), (r + 2), (r + 3)$ .
Consider the product: $(r + 1) (r + 2) (r + 3) = r^{3} + 6 r^{2} + 11 r + 6$ .
Compare this with our polynomial: $r^{3} + 6 r^{2} + 2 r + 5$ .

Adjusting the Terms

$(r + 1) (r + 2) (r + 3) = r^{3} + 6 r^{2} + 11 r + 6$
To get $r^{3} + 6 r^{2} + 2 r + 5$ , we subtract $9 r + 1$ from the product.
Rewrite $- (9 r + 1)$ as $- 9 (r + 1) + 8$ .
Final Decomposition: $r^{3} + 6 r^{2} + 2 r + 5 = (r + 1) (r + 2) (r + 3) - 9 (r + 1) + 8$ .

Forming Higher Factorials

Substitute the decomposition into the sum:
$\sum_{r = 1}^{10} r! {(r + 1) (r + 2) (r + 3) - 9 (r + 1) + 8}$
Distribute $r!$ :
$\sum_{r = 1}^{10} [(r + 3)! - 9 (r + 1)! + 8 r!]$

Linearity of Summation

Using linearity of $\sum$ :
$\sum_{r = 1}^{10} (r + 3)! - 9 \sum_{r = 1}^{10} (r + 1)! + 8 \sum_{r = 1}^{10} r!$
Let's denote these sums as $S_{1}, S_{2},$ and $S_{3}$ respectively.

Writing Out the Terms

$S_{1} = 4! + 5! + \dots + 10! + 11! + 12! + 13!$
$S_{2} = 2! + 3! + 4! + 5! + \dots + 10! + 11!$
$S_{3} = 1! + 2! + 3! + 4! + 5! + \dots + 10!$

Canceling Intermediate Terms

Total Sum $= S_{1} - 9 S_{2} + 8 S_{3}$
Coefficient of $k!$ for $4 \leq k \leq 10$ is $(1 - 9 + 8) = 0$ .
Remaining terms from $S_{1}$ : $11! + 12! + 13!$
Remaining terms from $S_{2}$ : $- 9 (2! + 3!) - 9 (11!)$
Remaining terms from $S_{3}$ : $8 (1! + 2! + 3!)$

Evaluating Lower Factorials

Constant part: $- 9 (2! + 3!) + 8 (1! + 2! + 3!)$
$= - 9 (2 + 6) + 8 (1 + 2 + 6)$
$= - 9 (8) + 8 (9) = - 72 + 72 = 0$
The entire sum simplifies to just the higher factorial terms: $13! + 12! + 11! - 9 (11!) = 13! + 12! - 8 (11!)$

Simplifying the Higher Terms

Sum $= 13! + 12! - 8 (11!)$
Factor out $11!$ :
$= (13 \times 12 \times 11!) + (12 \times 11!) - 8 (11!)$
$= (13 \times 12 + 12 - 8) \times 11!$

Finding $α$

Calculate the value inside the bracket:
$156 + 12 - 8 = 168 - 8 = 160$
So, the sum is $160 (11!)$ .
Comparing with $α (11!)$ , we get $α = 160$ .

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The Sigma Insight: Sum of Special Series

Analyzing the Setup

The problem asks us to evaluate the summation:

r = 1 \sum 10 r! (r^{3} + 6 r^{2} + 2 r + 5) = α (11!)

At first glance, this appears to be a complex arithmetic task. However, in the context of JEE Advanced, we look for hidden symmetry. When a factorial

r!

is multiplied by a polynomial, the intended strategy is almost always to construct a telescoping series.

The Algebraic Bridge

We aim to express the polynomial

P (r) = r^{3} + 6 r^{2} + 2 r + 5

in terms of consecutive factors like

(r + 1), (r + 2),

and

(r + 3)

. This is effective because

(r + 1) (r + 2) (r + 3) r! = (r + 3)!

First, consider the expansion:

(r + 1) (r + 2) (r + 3) = r^{3} + 6 r^{2} + 11 r + 6

Comparing this to our

P (r)

, we find the difference:

(r^{3} + 6 r^{2} + 11 r + 6) - (r^{3} + 6 r^{2} + 2 r + 5) = 9 r + 1

We can rewrite

9 r + 1

9 (r + 1) - 8

. Therefore, the decomposition of our polynomial is:

P (r) = (r + 1) (r + 2) (r + 3) - 9 (r + 1) + 8

The Telescoping Symphony

Substituting this decomposition back into the original summation, we get:

r = 1 \sum 10 [(r + 3)! - 9 (r + 1)! + 8 (r!)]

By the linearity of summation, we split this into three parts:

S = r = 1 \sum 10 (r + 3)! - 9 r = 1 \sum 10 (r + 1)! + 8 r = 1 \sum 10 r!

When expanding these terms, the values from

4!

10!

appear in all three sums with coefficients

1, - 9,

and

8

. Since

1 - 9 + 8 = 0

, these terms cancel out entirely.

We are left only with the boundary terms:

S = (4! + 5! + \dots + 13!) - 9 (2! + 3! + \dots + 11!) + 8 (1! + 2! + \dots + 10!)

After accounting for the cancellations, the expression simplifies to:

S = 13! + 12! - 8 (11!)

Final Calculation

We need the result in the form

α (11!)

. We factor out

11!

from the remaining terms:

13! = 13 \times 12 \times 11! = 156 (11!)

12! = 12 \times 11!

Substituting these back into the expression:

S = (156 + 12 - 8) \times 11!

S = 160 \times 11!

Thus, the value of

α

is 160.

Common Traps & FAQs

Why do we use

(r + 1) (r + 2) (r + 3)

? Why not just

(r + 1) (r + 2)

Great question! The degree of your polynomial is 3 (

r^{3}

). To 'absorb' a cubic term into a factorial, you need three consecutive factors to step up the factorial from

r!

(r + 3)!

. If the polynomial were quadratic, you would only need two. Always match the number of consecutive factors to the degree of the polynomial.

I'm confused about the cancellation. Do the limits of the summation change?

The limits don't change, but the terms do! When you write out the sums, you'll see that the terms from

r = 4

r = 10

appear in all three summations. Because of the coefficients (

1, - 9, 8

), they sum to zero. This is the 'telescoping' effect. Don't worry about the limits; just focus on the pattern of the terms.

Is there a faster way than writing out all the terms?

For the JEE, precision is speed. While you can use the formula for telescoping sums, writing out the first three and last three terms is the safest way to avoid sign errors. It takes 30 seconds and guarantees you won't miss a term. Trust the process!

Animated Solution for Mathematics - Sequence and Series: If $\sum_{r = 1}^{10} r! (r^{3} + 6 r^{2} + 2 r + 5) = α (11!)$ , then the value of $α$ is equal to

Enter Numerical Value:

Visualized Solution

$\sum_{r = 1}^{10} r! (r^{3} + 6 r^{2} + 2 r + 5) = α (11!)$

Expressing $P (r)$ in terms of $(r + 1), (r + 2), (r + 3)$

Adjusting the Terms

Forming Higher Factorials

Linearity of Summation

Writing Out the Terms

Canceling Intermediate Terms

Evaluating Lower Factorials

Simplifying the Higher Terms

Finding $α$

The Sigma Insight: Sum of Special Series

Analyzing the Setup

The Algebraic Bridge

The Telescoping Symphony

Final Calculation

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; } If ∑r=110​r!(r3+6r2+2r+5)=α(11!), then the value of α is equal to

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∑r=110​r!(r3+6r2+2r+5)=α(11!)

Expressing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(r) in terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (r+1),(r+2),(r+3)

Adjusting the Terms

Forming Higher Factorials

Linearity of Summation

Writing Out the Terms

Canceling Intermediate Terms

Evaluating Lower Factorials

Simplifying the Higher Terms

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

The Sigma Insight: Sum of Special Series

Analyzing the Setup

The Algebraic Bridge

The Telescoping Symphony

Final Calculation

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Animated Solution for Mathematics - Sequence and Series: If $\sum_{r = 1}^{10} r! (r^{3} + 6 r^{2} + 2 r + 5) = α (11!)$ , then the value of $α$ is equal to

$\sum_{r = 1}^{10} r! (r^{3} + 6 r^{2} + 2 r + 5) = α (11!)$

Expressing $P (r)$ in terms of $(r + 1), (r + 2), (r + 3)$

Finding $α$

The value of $lim_{n \to \infty} (\sum_{k = 1}^{n} \frac{k ^{3} + 6 k ^{2} + 11 k + 5}{( k + 3 )!})$ is equal to:

The value of $\sum_{k = 1}^{\infty} (- 1)^{k + 1} \frac{k ( k + 1 )}{k !}$ is

$\sum_{r = 1}^{20} (r^{2} + 1) (r!)$ is equal to:

If $(\frac{1}{α + 1} + \frac{1}{α + 2} + \dots + \frac{1}{α + 1012}) - (\frac{1}{2 \cdot 1} + \frac{1}{4 \cdot 3} + \frac{1}{6 \cdot 5} + \dots + \frac{1}{2024 \cdot 2023}) = \frac{1}{2024}$ , then $α$ is equal to

If $\sum_{r = 1}^{n} T_{r} = \frac{( 2 n - 1 ) ( 2 n + 1 ) ( 2 n + 3 ) ( 2 n + 5 )}{64}$ then $lim_{n \to \infty} \sum_{r = 1}^{n} (\frac{1}{T _{r}})$ is equal to:

The sum of the series $\sum_{n = 1}^{\infty} \frac{n ^{2} + 6 n + 10}{( 2 n + 1 )!}$ is equal to :

The sum $1 + \frac{1 + 3}{2 !} + \frac{1 + 3 + 5}{3 !} + \frac{1 + 3 + 5 + 7}{4 !} + \dots$ upto $\infty$ terms, is equal to

The sum $\sum_{k = 1}^{20} (1 + 2 + 3 + .... + k)$ is _____________.

For $k \in N$ , if the sum of the series $1 + \frac{4}{k} + \frac{8}{k ^{2}} + \frac{13}{k ^{3}} + \frac{19}{k ^{4}} + \dots$ is 10, then the value of $k$ is

For positive integers n, if $4 a_{n} = (n^{2} + 5 n + 6)$ and $S_{n} = \sum_{k = 1}^{n} (\frac{1}{a _{k}})$ , then the value of $507 S_{2025}$ is: