Is there a faster way to get $b_n = \alpha^n + \beta^n$?

Yes, you could use Newton's Sums, which relate the sums of powers of roots to the coefficients of the polynomial. However, the algebraic manipulation we used—replacing $1$ with $-\alpha\beta$—is a classic JEE technique that builds deep intuition for how roots interact with their own equations. It is worth mastering this method!

JEE Advanced 2019

LEVELJEE Advanced

Animated Solution for Mathematics - Sequence and Series: Let $α$ and $β$ be the roots of $x^{2} - x - 1 = 0$ , with $α > β$ . For all positive integers $n$ , define $a_{n} = \frac{α ^{n} - β ^{n}}{α - β}, n \geq 1$ , $b_{1} = 1$ and $b_{n} = a_{n - 1} + a_{n + 1}, n \geq 2$ . Then which of the following options is/are correct?

Q: How did you know to use generating functions for the infinite sum?

Whenever you see an infinite sum of the form $\sum \frac{a_n}{k^n}$, it is a massive hint to look at the power series $S(x) = \sum a_n x^n$ evaluated at $x = \frac{1}{k}$. It transforms a difficult summation problem into a simple algebraic equation. It is a powerful tool in your JEE arsenal!

Select Answer:

* Multiple Correct

Progressive Hints

Hint 1

Hint 1: The Spark. The quadratic equation

x^{2} - x - 1 = 0

is the characteristic equation for the Fibonacci sequence. Notice that if

α

and

β

are roots, then

α^{2} = α + 1

and

β^{2} = β + 1

. This implies that any sequence defined by

a_{n} = c_{1} α^{n} + c_{2} β^{n}

will satisfy the recurrence relation

a_{n + 2} = a_{n + 1} + a_{n}

Hint 2

Hint 2: The Arsenal. To handle the infinite sum

\sum_{n = 1}^{\infty} \frac{a _{n}}{1 0 ^{n}}

, consider the generating function

S (x) = \sum_{n = 1}^{\infty} a_{n} x^{n}

. Use the recurrence relation

a_{n} = a_{n - 1} + a_{n - 2}

to express

S (x)

in terms of itself, allowing you to solve for

S (x)

algebraically.

Hint 3

Hint 3: The Execution. For the sequence

b_{n} = a_{n - 1} + a_{n + 1}

, do not expand the powers directly. Instead, use the property

1 = - α β

to rewrite

1 + α^{2}

α (α - β)

. This will allow you to simplify the expression for

b_{n}

into the elegant form

b_{n} = α^{n} + β^{n}

Visualized Solution

Roots of the Quadratic

Given equation: $x^{2} - x - 1 = 0$
Sum of roots: $α + β = 1$
Product of roots: $α β = - 1$
Since $α, β$ are roots, they satisfy the equation:
$α^{2} = α + 1$ and $β^{2} = β + 1$

Recurrence Relation for $a_{n}$

Given: $a_{n} = \frac{α ^{n} - β ^{n}}{α - β}$
Using $α^{2} = α + 1$ , multiply by $α^{n}$ :
$α^{n + 2} = α^{n + 1} + α^{n}$
Similarly: $β^{n + 2} = β^{n + 1} + β^{n}$
Subtracting the two equations and dividing by $(α - β)$ gives:
$a_{n + 2} = a_{n + 1} + a_{n}$

Evaluating Option A

From recurrence: $a_{n} = a_{n + 2} - a_{n + 1}$
Write terms vertically:
$a_{1} = a_{3} - a_{2}$
$a_{2} = a_{4} - a_{3}$
$\dots$
$a_{n} = a_{n + 2} - a_{n + 1}$
Summing vertically (telescoping sum):
$\sum_{i = 1}^{n} a_{i} = a_{n + 2} - a_{2}$
Since $a_{2} = \frac{α ^{2} - β ^{2}}{α - β} = α + β = 1$ :
$\sum_{i = 1}^{n} a_{i} = a_{n + 2} - 1$ (Option A is correct)

Infinite Sum of $a_{n}$

Let $S = \sum_{n = 1}^{\infty} a_{n} x^{n} = a_{1} x + a_{2} x^{2} + \sum_{n = 3}^{\infty} a_{n} x^{n}$
Substitute $a_{n} = a_{n - 1} + a_{n - 2}$ for $n \geq 3$ :
$S = x + x^{2} + \sum_{n = 3}^{\infty} (a_{n - 1} + a_{n - 2}) x^{n}$
Split the summation:
$S = x + x^{2} + x \sum_{n = 3}^{\infty} a_{n - 1} x^{n - 1} + x^{2} \sum_{n = 3}^{\infty} a_{n - 2} x^{n - 2}$

Solving the Infinite Sum

Notice the shifted indices in the summations:
$S = x + x^{2} + x (S - a_{1} x) + x^{2} S$
Since $a_{1} = 1$ , substitute and expand:
$S = x + x^{2} + x S - x^{2} + x^{2} S$
The $x^{2}$ terms cancel out:
$S = x + x S + x^{2} S$
Rearrange to solve for S:
$S (1 - x - x^{2}) = x ⟹ S = \frac{x}{1 - x - x ^{2}}$

Evaluating Option B

We need $\sum_{n = 1}^{\infty} \frac{a _{n}}{1 0 ^{n}}$ , which is $S$ evaluated at $x = \frac{1}{10}$
Substitute $x = \frac{1}{10}$ into the generating function:
$S = \frac{1/10}{1 - 1/10 - 1/100}$
$S = \frac{1/10}{( 100 - 10 - 1 ) /100}$
$S = \frac{1/10}{89/100} = \frac{10}{89}$
Option B is correct.

Simplifying $b_{n}$

Given $b_{n} = a_{n - 1} + a_{n + 1}$
Substitute the definition of $a_{n}$ :
$b_{n} = \frac{α ^{n - 1} - β ^{n - 1} + α ^{n + 1} - β ^{n + 1}}{α - β}$
Group the $α$ terms and $β$ terms:
$b_{n} = \frac{α ^{n - 1} ( 1 + α ^{2} ) - β ^{n - 1} ( 1 + β ^{2} )}{α - β}$

A Beautiful Algebraic Trick

We need to simplify the term $(1 + α^{2})$ .
Recall that the product of roots is $α β = - 1 ⟹ 1 = - α β$
Substitute this into the expression:
$1 + α^{2} = - α β + α^{2} = α (α - β)$
Similarly, apply this to the $β$ term:
$1 + β^{2} = - α β + β^{2} = β (β - α) = - β (α - β)$

Evaluating Option D

Substitute the simplified terms back into $b_{n}$ :
$b_{n} = \frac{α ^{n - 1} ( α ( α - β )) - β ^{n - 1} ( - β ( α - β ))}{α - β}$
Combine the powers of $α$ and $β$ :
$b_{n} = \frac{α ^{n} ( α - β ) + β ^{n} ( α - β )}{α - β}$
Factor out $(α - β)$ and cancel with the denominator:
$b_{n} = α^{n} + β^{n}$ (Option D is correct)

Infinite Sum of $b_{n}$

Let $S^{'} = \sum_{n = 1}^{\infty} b_{n} x^{n} = \sum_{n = 1}^{\infty} (α^{n} + β^{n}) x^{n}$
Split into two infinite geometric progressions:
$S^{'} = \sum_{n = 1}^{\infty} (α x)^{n} + \sum_{n = 1}^{\infty} (β x)^{n}$
Using the sum of an infinite GP ( $S = \frac{a}{1 - r}$ ):
$S^{'} = \frac{α x}{1 - α x} + \frac{β x}{1 - β x}$

Simplifying the Sum

Take a common denominator and cross-multiply:
$S^{'} = \frac{α x ( 1 - β x ) + β x ( 1 - α x )}{( 1 - α x ) ( 1 - β x )}$
Expand the numerator and denominator:
$S^{'} = \frac{( α + β ) x - 2 α β x ^{2}}{1 - ( α + β ) x + α β x ^{2}}$
Substitute $α + β = 1$ and $α β = - 1$ :
$S^{'} = \frac{x - 2 ( - 1 ) x ^{2}}{1 - ( 1 ) x + ( - 1 ) x ^{2}}$
$S^{'} = \frac{x + 2 x ^{2}}{1 - x - x ^{2}}$

Evaluating Option C

We need $\sum_{n = 1}^{\infty} \frac{b _{n}}{1 0 ^{n}}$ , which is $S^{'}$ evaluated at $x = \frac{1}{10}$
Substitute $x = \frac{1}{10}$ into the simplified $S^{'}$ :
$S^{'} = \frac{1/10 + 2/100}{1 - 1/10 - 1/100}$
$S^{'} = \frac{12/100}{89/100} = \frac{12}{89}$
Option C claims the sum is $\frac{8}{89}$ , which is incorrect.
Final Conclusion: Options A, B, and D are correct.

00:00 / 00:00

The Sigma Insight: Sum of Special Series

Solution Diagram

Analyzing the Setup

The quadratic equation

x^{2} - x - 1 = 0

serves as the foundation for this mathematical landscape. The roots,

α

and

β

, are governed by Vieta's formulas:

α + β = 1

and

α β = - 1

Because

α

and

β

are roots of the quadratic, they satisfy the following identities:

α^{2} = α + 1

β^{2} = β + 1

These identities act as our master key for manipulating higher powers of the roots.

The Recurrence Relation

We define the sequence

a_{n}

as:

a_{n} = \frac{α ^{n} - β ^{n}}{α - β}

By multiplying the identity

α^{2} = α + 1

α^{n}

, we obtain

α^{n + 2} = α^{n + 1} + α^{n}

. Applying the same logic to

β

and subtracting the two resulting equations yields the Fibonacci recurrence:

a_{n + 2} = a_{n + 1} + a_{n}

This confirms that every term in the sequence is the sum of the two preceding terms.

The Telescoping Magic

To evaluate the sum

\sum_{i = 1}^{n} a_{i}

, we rearrange the recurrence relation into

a_{i} = a_{i + 2} - a_{i + 1}

. Writing out the terms:

a_{1} = a_{3} - a_{2}

a_{2} = a_{4} - a_{3}

...

a_{n} = a_{n + 2} - a_{n + 1}

Summing these vertically results in a telescoping sum where intermediate terms cancel out. We are left with:

i = 1 \sum n a_{i} = a_{n + 2} - a_{2}

Since

a_{2} = \frac{α ^{2} - β ^{2}}{α - β} = α + β = 1

, the sum simplifies to

a_{n + 2} - 1

. Thus, Option A is confirmed.

The Power of Generating Functions

To evaluate the infinite sum

\sum_{n = 1}^{\infty} \frac{a _{n}}{1 0 ^{n}}

, we define the generating function

S = \sum_{n = 1}^{\infty} a_{n} x^{n}

. Using the recurrence

a_{n} = a_{n - 1} + a_{n - 2}

, we derive:

S = \frac{x}{1 - x - x ^{2}}

Evaluating this at

x = \frac{1}{10}

yields:

S = \frac{1/10}{1 - 1/10 - 1/100} = \frac{10}{89}

This confirms that Option B is correct.

The Algebraic Beauty of $b_{n}$

We examine the sequence

b_{n} = a_{n - 1} + a_{n + 1}

. By substituting the definition of

a_{n}

and utilizing the identity

1 = - α β

, we simplify the expression:

b_{n} = α^{n} + β^{n}

This elegant result confirms that Option D is correct. Conversely, evaluating the sum for

b_{n}

x = \frac{1}{10}

results in

\frac{12}{89}

, which proves that Option C is incorrect.

Common Traps & FAQs

Wait, why can we just assume

a_{n} = a_{n - 1} + a_{n - 2}

? Is it always true for these types of sequences?

That is a fantastic question! It is not an assumption; it is a direct consequence of the roots satisfying the quadratic equation. Since

α^{2} = α + 1

, multiplying by

α^{n - 1}

gives

α^{n + 1} = α^{n} + α^{n - 1}

. The same holds for

β

. When you take the linear combination

a_{n} = \frac{α ^{n} - β ^{n}}{α - β}

, the recurrence relation is inherited from the roots. It is the fundamental heartbeat of the Fibonacci sequence!

How did you know to use generating functions for the infinite sum?

Whenever you see an infinite sum of the form

\sum \frac{a _{n}}{k ^{n}}

, it is a massive hint to look at the power series

S (x) = \sum a_{n} x^{n}

evaluated at

x = \frac{1}{k}

. It transforms a difficult summation problem into a simple algebraic equation. It is a powerful tool in your JEE arsenal!

Is there a faster way to get

b_{n} = α^{n} + β^{n}

Yes, you could use Newton's Sums, which relate the sums of powers of roots to the coefficients of the polynomial. However, the algebraic manipulation we used—replacing

1

with

- α β

—is a classic JEE technique that builds deep intuition for how roots interact with their own equations. It is worth mastering this method!

Select Answer:

Visualized Solution

Roots of the Quadratic

Recurrence Relation for $a_{n}$

Evaluating Option A

Infinite Sum of $a_{n}$

Solving the Infinite Sum

Evaluating Option B

Simplifying $b_{n}$

A Beautiful Algebraic Trick

Evaluating Option D

Infinite Sum of $b_{n}$

Simplifying the Sum

Evaluating Option C

The Sigma Insight: Sum of Special Series

Analyzing the Setup

The Recurrence Relation

The Telescoping Magic

The Power of Generating Functions

The Algebraic Beauty of $b_{n}$

Similar Questions

For a positive integer $n$ , let $a (n) = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots + \frac{1}{( 2 ^{n} ) - 1}$ . Then

If $a_{n} = \frac{- 2}{4 n ^{2} - 16 n + 15}$ , then $a_{1} + a_{2} + \dots + a_{25}$ is equal to :

Statement-1 : For every natural number $n \geq 2, \frac{1}{1} + \frac{1}{2} + \dots + \frac{1}{n} > n$ . Statement-2 : For every natural number $n \geq 2, n (n + 1) < n + 1$ .

If $a_{n} = 7 + 7 + 7 + ....$ having $n$ radical signs then by methods of mathematical induction which is true

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Select Answer:

Visualized Solution

Roots of the Quadratic

Recurrence Relation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } an​

Evaluating Option A

Infinite Sum of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } an​

Solving the Infinite Sum

Evaluating Option B

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

A Beautiful Algebraic Trick

Evaluating Option D

Infinite Sum of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } bn​

Simplifying the Sum

Evaluating Option C

The Sigma Insight: Sum of Special Series

Analyzing the Setup

The Recurrence Relation

The Telescoping Magic

The Power of Generating Functions

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

Similar Questions

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Recurrence Relation for $a_{n}$

Infinite Sum of $a_{n}$

Simplifying $b_{n}$

Infinite Sum of $b_{n}$

The Algebraic Beauty of $b_{n}$

For a positive integer $n$ , let $a (n) = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots + \frac{1}{( 2 ^{n} ) - 1}$ . Then

If $a_{n} = \frac{- 2}{4 n ^{2} - 16 n + 15}$ , then $a_{1} + a_{2} + \dots + a_{25}$ is equal to :

Statement-1 : For every natural number $n \geq 2, \frac{1}{1} + \frac{1}{2} + \dots + \frac{1}{n} > n$ . Statement-2 : For every natural number $n \geq 2, n (n + 1) < n + 1$ .

If $a_{n} = 7 + 7 + 7 + ....$ having $n$ radical signs then by methods of mathematical induction which is true

Sum of the first $n$ terms of the series $\frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{15}{16} + \dots$ is equal to

Let $S$ be the sum of the first 9 terms of the series: $x + k a + x^{2} + (k + 2) a + x^{3} + (k + 4) a + x^{4} + (k + 6) a + ...$ where $a \neq = 0$ and $x \neq = 1$ If $S = \frac{x ^{10} - x + 45 a ( x - 1 )}{x - 1}$ , then $k$ is equal to :

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Let $α = 1^{2} + 4^{2} + 8^{2} + 1 3^{2} + 1 9^{2} + 2 6^{2} + \dots$ upto 10 terms and $β = \sum_{n = 1}^{10} n^{4}$ . If $4 α - β = 55 k + 40$ , then $k$ is equal to

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