JEE Advanced 2010

LEVELJEE Main

Animated Solution for Mathematics - Sequence and Series: Let $S_{k}, k = 1, 2, \dots, 100$ , denote the sum of the infinite geometric series whose first term is $\frac{k - 1}{k !}$ and the common ratio is $\frac{1}{k}$ . Then the value of $\frac{10 0 ^{2}}{100 !} + \sum_{k = 1}^{100} ∣ (k^{2} - 3 k + 1) S_{k} ∣$ is \dots.

Q: Wait, why is $S_1 = 0$? The formula $S = a/(1-r)$ gives $0/0$.

That is a fantastic observation! The formula $S = a/(1-r)$ is only valid when $|r| < 1$. For $k=1$, $r=1$, so the formula fails. Instead, look at the series itself: the first term is $(1-1)/1! = 0$, and the common ratio is $1$. The series is $0 + 0 + 0 + \dots$, which clearly sums to $0$. Always check the validity of your formulas before applying them!

Enter Numerical Value:

Progressive Hints

Hint 1

The Spark: Before diving into the algebra, look closely at the definition of

S_{k}

. For

k = 1

, the common ratio

r = 1/1 = 1

. Does the infinite geometric series formula

S = a / (1 - r)

apply here? Visualize the series for

k = 1

and see if it behaves like a standard G.P.

Hint 2

The Arsenal: Once you have

S_{k} = 1/ (k - 1)!

for

k \geq 2

, you face the term

∣ (k^{2} - 3 k + 1) S_{k} ∣

. To handle the absolute value, test the sign of the quadratic

k^{2} - 3 k + 1

for

k \geq 3

. You will find it is always positive, allowing you to remove the absolute value bars.

Hint 3

The Execution: To solve the summation, manipulate the numerator

k^{2} - 3 k + 1

to create a telescoping structure. Rewrite it as

(k - 1) (k - 2) - 1

. This will allow you to split the term into two parts:

1/ (k - 3)! - 1/ (k - 1)!

. Now, write out the first few terms and watch the magic of cancellation unfold.

Visualized Solution

Identifying the Infinite G.P.

Given an infinite geometric series for each $k$ .
First term: $a = \frac{k - 1}{k !}$
Common ratio: $r = \frac{1}{k}$

Formula for Infinite G.P. Sum

Sum of an infinite G.P.: $S = \frac{a}{1 - r}$
Valid when $∣ r ∣ < 1$ .
Here, $r = \frac{1}{k}$ , so $∣ r ∣ < 1$ for $k \geq 2$ .

Substituting Values into the Sum

For $k \geq 2$ , substitute $a$ and $r$ into the sum formula.
$S_{k} = \frac{\frac{k - 1}{k !}}{1 - \frac{1}{k}}$

Simplifying the Denominator

Simplify the denominator: $1 - \frac{1}{k} = \frac{k - 1}{k}$
$S_{k} = \frac{\frac{k - 1}{k !}}{\frac{k - 1}{k}}$

Simplifying $S_{k}$

Cancel $(k - 1)$ from numerator and denominator.
$S_{k} = \frac{k}{k !} = \frac{k}{k \cdot ( k - 1 )!}$
$S_{k} = \frac{1}{( k - 1 )!}$ for $k \geq 2$

Handling the Boundary Case $k = 1$

What happens when $k = 1$ ?
First term $a = \frac{1 - 1}{1 !} = 0$
Common ratio $r = \frac{1}{1} = 1$
Since $a = 0$ , the series is $0 + 0 + 0 + \dots$
Therefore, $S_{1} = 0$

Defining the Summation Term $T_{k}$

Let the term inside the summation be $T_{k} = ∣ (k^{2} - 3 k + 1) S_{k} ∣$
Evaluate for $k = 1$ : $T_{1} = ∣ (1 - 3 + 1) \cdot 0∣ = 0$
Evaluate for $k = 2$ : $T_{2} = ∣ (4 - 6 + 1) \cdot S_{2} ∣ = ∣ - 1 \cdot 1∣ = 1$

Analyzing the Sign for $k \geq 3$

For $k \geq 3$ , analyze the quadratic $k^{2} - 3 k + 1$
$k^{2} - 3 k + 1 = k (k - 3) + 1$
Since $k \geq 3$ , $k (k - 3) \geq 0$ , so $k^{2} - 3 k + 1 > 0$
Therefore, the absolute value can be removed: $T_{k} = (k^{2} - 3 k + 1) S_{k}$

Substituting $S_{k}$ into $T_{k}$

For $k \geq 3$ , substitute $S_{k} = \frac{1}{( k - 1 )!}$
$T_{k} = \frac{k ^{2} - 3 k + 1}{( k - 1 )!}$
We need to manipulate the numerator to match the factorial in the denominator.

Manipulating the Numerator

Express $k^{2} - 3 k + 1$ in terms of $(k - 1)$ and $(k - 2)$
$k^{2} - 3 k + 1 = k^{2} - 3 k + 2 - 1$
Factorize: $(k - 1) (k - 2) - 1$
So, $T_{k} = \frac{( k - 1 ) ( k - 2 ) - 1}{( k - 1 )!}$

Splitting $T_{k}$ into Two Terms

Split the fraction: $T_{k} = \frac{( k - 1 ) ( k - 2 )}{( k - 1 )!} - \frac{1}{( k - 1 )!}$
Simplify the first term: $\frac{( k - 1 ) ( k - 2 )}{( k - 1 ) ( k - 2 ) ( k - 3 )!} = \frac{1}{( k - 3 )!}$
$T_{k} = \frac{1}{( k - 3 )!} - \frac{1}{( k - 1 )!}$

Setting up the Telescoping Sum

We need to sum $T_{k}$ from $k = 3$ to $100$ .
$\sum_{k = 3}^{100} T_{k} = \sum_{k = 3}^{100} (\frac{1}{( k - 3 )!} - \frac{1}{( k - 1 )!})$
Write out the first few terms:
$k = 3 : \frac{1}{0 !} - \frac{1}{2 !}$
$k = 4 : \frac{1}{1 !} - \frac{1}{3 !}$
$k = 5 : \frac{1}{2 !} - \frac{1}{4 !}$

Evaluating the Telescoping Sum

Observe the cancellations: $- \frac{1}{2 !}$ cancels with $+ \frac{1}{2 !}$ , etc.
All terms cancel except the first two positive terms and the last two negative terms.
Positive survivors (from $k = 3, 4$ ): $\frac{1}{0 !} + \frac{1}{1 !} = 1 + 1 = 2$
Negative survivors (from $k = 99, 100$ ): $- \frac{1}{98 !} - \frac{1}{99 !}$
Sum for $k \geq 3$ : $2 - \frac{1}{98 !} - \frac{1}{99 !}$

Total Summation $\sum_{k = 1}^{100} T_{k}$

Add the terms for $k = 1$ and $k = 2$ back into the total sum.
$\sum_{k = 1}^{100} T_{k} = T_{1} + T_{2} + \sum_{k = 3}^{100} T_{k}$
$\sum_{k = 1}^{100} T_{k} = 0 + 1 + (2 - \frac{1}{98 !} - \frac{1}{99 !})$
$\sum_{k = 1}^{100} T_{k} = 3 - \frac{1}{98 !} - \frac{1}{99 !}$

Final Expression Evaluation

The problem asks for: $\frac{10 0 ^{2}}{100 !} + \sum_{k = 1}^{100} T_{k}$
Substitute the summation: $\frac{10 0 ^{2}}{100 !} + 3 - \frac{1}{98 !} - \frac{1}{99 !}$
Simplify the first term: $\frac{100 \cdot 100}{100 \cdot 99 !} = \frac{100}{99 !} = \frac{99 + 1}{99 !} = \frac{99}{99 !} + \frac{1}{99 !} = \frac{1}{98 !} + \frac{1}{99 !}$
Final result: $(\frac{1}{98 !} + \frac{1}{99 !}) + 3 - \frac{1}{98 !} - \frac{1}{99 !} = 3$

00:00 / 00:00

The Sigma Insight: Sum of Special Series

Analyzing the Series Definition

We are tasked with evaluating the sum of an infinite geometric series where the first term is

a = \frac{k - 1}{k !}

and the common ratio is

r = \frac{1}{k}

. The sum of an infinite geometric progression is given by the formula:

S = \frac{a}{1 - r}

This formula is valid provided that

∣ r ∣ < 1

. For the case

k = 1

, we observe that

r = 1

. However, since the first term

a = \frac{1 - 1}{1 !} = 0

, the series becomes

0 + 0 + 0...

, yielding

S_{1} = 0

Deriving the General Term

For

k \geq 2

, the condition

∣ r ∣ < 1

is satisfied. Substituting our values for

a

and

r

into the sum formula, we obtain:

S_{k} = \frac{\frac{k - 1}{k !}}{1 - \frac{1}{k}} = \frac{\frac{k - 1}{k !}}{\frac{k - 1}{k}}

Simplifying this expression, we find:

S_{k} = \frac{k - 1}{k !} \cdot \frac{k}{k - 1} = \frac{k}{k !} = \frac{1}{( k - 1 )!}

Evaluating the Summation

We now focus on the summation

\sum_{k = 1}^{100} ∣ (k^{2} - 3 k + 1) S_{k} ∣

. For

k = 1

, the term is

0

. For

k = 2

, the term is

∣ (4 - 6 + 1) \cdot 1∣ = 1

For

k \geq 3

, the quadratic expression

k^{2} - 3 k + 1

is strictly positive. We can rewrite the numerator to facilitate a telescoping series:

k^{2} - 3 k + 1 = (k^{2} - 3 k + 2) - 1 = (k - 1) (k - 2) - 1

Substituting this into our summation term, we get:

\frac{( k - 1 ) ( k - 2 ) - 1}{( k - 1 )!} = \frac{( k - 1 ) ( k - 2 )}{( k - 1 ) ( k - 2 ) ( k - 3 )!} - \frac{1}{( k - 1 )!} = \frac{1}{( k - 3 )!} - \frac{1}{( k - 1 )!}

Final Calculation

The summation now takes the form of a telescoping series from

k = 3

100

. As the terms cancel out, we are left with the boundary values:

k = 3 \sum 100 (\frac{1}{( k - 3 )!} - \frac{1}{( k - 1 )!}) = (\frac{1}{0 !} + \frac{1}{1 !}) - (\frac{1}{98 !} + \frac{1}{99 !})

Adding the previously calculated term for

k = 2

(which is

1

) and accounting for the remaining terms, the factorial components effectively resolve. The final result of this summation is:

Common Traps & FAQs

Wait, why is

S_{1} = 0

? The formula

S = a / (1 - r)

gives

0/0

That is a fantastic observation! The formula

S = a / (1 - r)

is only valid when

∣ r ∣ < 1

. For

k = 1

r = 1

, so the formula fails. Instead, look at the series itself: the first term is

(1 - 1) /1! = 0

, and the common ratio is

1

. The series is

0 + 0 + 0 + \dots

, which clearly sums to

0

. Always check the validity of your formulas before applying them!

Why can we just drop the absolute value for

k \geq 3

Great question. We need to ensure the expression inside the absolute value,

k^{2} - 3 k + 1

, is non-negative. For

k = 3

9 - 9 + 1 = 1 > 0

. For

k = 4

16 - 12 + 1 = 5 > 0

. As

k

increases, the quadratic

k^{2} - 3 k + 1

grows. Since the vertex is at

k = 1.5

, for all

k \geq 3

, the function is strictly increasing and positive. Thus,

∣ k^{2} - 3 k + 1∣ = k^{2} - 3 k + 1

How did you know to rewrite

k^{2} - 3 k + 1

(k - 1) (k - 2) - 1

This is a classic JEE strategy! We look at the denominator, which is

(k - 1)!

. We want to cancel terms to simplify the factorial. By expressing the numerator in terms of

(k - 1)

and

(k - 2)

, we can divide by

(k - 1)!

and get terms like

1/ (k - 3)!

. It is all about making the numerator 'match' the denominator's factorial structure.

Enter Numerical Value:

Visualized Solution

Identifying the Infinite G.P.

Formula for Infinite G.P. Sum

Substituting Values into the Sum

Simplifying the Denominator

Simplifying $S_{k}$

Handling the Boundary Case $k = 1$

Defining the Summation Term $T_{k}$

Analyzing the Sign for $k \geq 3$

Substituting $S_{k}$ into $T_{k}$

Manipulating the Numerator

Splitting $T_{k}$ into Two Terms

Setting up the Telescoping Sum

Evaluating the Telescoping Sum

Total Summation $\sum_{k = 1}^{100} T_{k}$

Final Expression Evaluation

The Sigma Insight: Sum of Special Series

Analyzing the Series Definition

Deriving the General Term

Evaluating the Summation

Final Calculation

Similar Questions

Let $⟨ a_{n} ⟩$ be a sequence such that $a_{0} = 0$ , $a_{1} = \frac{1}{2}$ and $2 a_{n + 2} = 5 a_{n + 1} - 3 a_{n}$ , $n = 0, 1, 2, 3, ...$ Then $\sum_{k = 1}^{100} a_{k}$ is

The sum of series $\frac{1}{2 !} - \frac{1}{3 !} + \frac{1}{4 !} - \dots$ upto infinity is

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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

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Let ${a_{n}}_{n = 1}^{\infty}$ be a sequence such that $a_{1} = 1, a_{2} = 1$ and $a_{n + 2} = 2 a_{n + 1} + a_{n}$ for all $n \geq 1$ . Then the value of $47 \sum_{n = 1}^{\infty} \frac{a _{n}}{2 ^{3 n}}$ is equal to

Enter Numerical Value:

Visualized Solution

Identifying the Infinite G.P.

Formula for Infinite G.P. Sum

Substituting Values into the Sum

Simplifying the Denominator

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

Handling the Boundary Case .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k=1

Defining the Summation Term .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tk​

Analyzing the Sign for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k≥3

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Sk​ into .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tk​

Manipulating the Numerator

Splitting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tk​ into Two Terms

Setting up the Telescoping Sum

Evaluating the Telescoping Sum

Total Summation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∑k=1100​Tk​

Final Expression Evaluation

The Sigma Insight: Sum of Special Series

Analyzing the Series Definition

Deriving the General Term

Evaluating the Summation

Final Calculation

Similar Questions

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Simplifying $S_{k}$

Handling the Boundary Case $k = 1$

Defining the Summation Term $T_{k}$

Analyzing the Sign for $k \geq 3$

Substituting $S_{k}$ into $T_{k}$

Splitting $T_{k}$ into Two Terms

Total Summation $\sum_{k = 1}^{100} T_{k}$

Let $⟨ a_{n} ⟩$ be a sequence such that $a_{0} = 0$ , $a_{1} = \frac{1}{2}$ and $2 a_{n + 2} = 5 a_{n + 1} - 3 a_{n}$ , $n = 0, 1, 2, 3, ...$ Then $\sum_{k = 1}^{100} a_{k}$ is

The sum of series $\frac{1}{2 !} - \frac{1}{3 !} + \frac{1}{4 !} - \dots$ upto infinity is

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The sum of the series $1 + \frac{1}{4.2 !} + \frac{1}{16.4 !} + \frac{1}{64.6 !} + \dots$ ad inf. is

The sum of $10$ terms of the series $\frac{3}{1 ^{2} \times 2 ^{2}} + \frac{5}{2 ^{2} \times 3 ^{2}} + \frac{7}{3 ^{2} \times 4 ^{2}} + \dots$ is :

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

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Let ${a_{n}}_{n = 1}^{\infty}$ be a sequence such that $a_{1} = 1, a_{2} = 1$ and $a_{n + 2} = 2 a_{n + 1} + a_{n}$ for all $n \geq 1$ . Then the value of $47 \sum_{n = 1}^{\infty} \frac{a _{n}}{2 ^{3 n}}$ is equal to