JEE Advanced 1992

LEVELJEE Main

Animated Solution for Mathematics - Sequence and Series: Find the natural number 'a' for which $\sum_{k = 1}^{n} f (a + k) = 16 (2^{n} - 1)$ , where the function 'f' satisfies the relation $f (x + y) = f (x) f (y)$ for all natural numbers $x, y$ and further $f (1) = 2$ .

Q: Why can we just cancel $(2^n - 1)$ from both sides?

We can cancel it because $n$ is a natural number ($n \geq 1$). For any $n \geq 1$, $2^n$ is at least 2, so $2^n - 1$ is at least 1. Since it is never zero, dividing both sides by this term is mathematically sound and perfectly safe.

Q: Is there a risk that $f(x)$ could be something other than $2^x$?

For natural numbers, the definition $f(x+y) = f(x)f(y)$ with $f(1)=2$ uniquely defines the function as $2^x$. If the domain were real numbers, we would need to assume continuity to rule out pathological solutions, but for natural numbers, the exponential form is the only one that satisfies the condition.

Q: Why did we pull $2^a$ out of the summation?

The summation operator $\sum$ acts only on the index $k$. Any term that does not contain $k$ is treated as a constant relative to the summation. Since $2^{a+k} = 2^a \cdot 2^k$, the $2^a$ is just a constant multiplier that can be moved to the front, simplifying the expression significantly.

Enter Numerical Value:

Visualized Solution

The Functional Equation

Given relation: $f (x + y) = f (x) f (y)$
This is a standard functional equation.
It represents an exponential function.

Deducing the Function Form

General form for $f (x + y) = f (x) f (y)$ is $f (x) = b^{x}$ .
Here, $b$ is a constant base.
We need a boundary condition to find $b$ .

Using the Boundary Condition

Given condition: $f (1) = 2$
Substitute $x = 1$ into our general form $f (x) = b^{x}$ .
$f (1) = b^{1}$

Finding the Base $b$

Equating the values: $b^{1} = 2$
Therefore, $b = 2$ .
The exact function is $f (x) = 2^{x}$ .

The Summation Equation

We need to evaluate: $\sum_{k = 1}^{n} f (a + k)$
This sum is given to be equal to $16 (2^{n} - 1)$ .

Substituting $f (x)$ into the Sum

Replace $f (a + k)$ using $f (x) = 2^{x}$ .
$f (a + k) = 2^{a + k}$
The sum becomes: $\sum_{k = 1}^{n} 2^{a + k}$

Splitting the Exponent

Using laws of exponents: $2^{a + k} = 2^{a} \cdot 2^{k}$
The term $2^{a}$ is independent of the summation index $k$ .

Factoring Out the Constant

Pull $2^{a}$ outside the summation.
Expression becomes: $2^{a} \sum_{k = 1}^{n} 2^{k}$
This equals $16 (2^{n} - 1)$ .

Identifying the Geometric Progression

Expand the sum: $\sum_{k = 1}^{n} 2^{k} = 2^{1} + 2^{2} + 2^{3} + \dots + 2^{n}$
This is a Geometric Progression (G.P.).
First term $a^{'} = 2$ , Common ratio $r = 2$ .

Applying the G.P. Sum Formula

G.P. Sum Formula: $S_{n} = \frac{a ^{'} ( r ^{n} - 1 )}{r - 1}$
Substitute $a^{'} = 2$ and $r = 2$ :
$S_{n} = \frac{2 ( 2 ^{n} - 1 )}{2 - 1}$

Simplifying the G.P. Sum

Denominator: $2 - 1 = 1$
The sum simplifies to: $2 (2^{n} - 1)$
Multiply with the constant outside: $2^{a} \cdot 2 (2^{n} - 1)$
Combine powers of 2: $2^{a + 1} (2^{n} - 1)$

Equating and Canceling Terms

Equate to the given RHS: $2^{a + 1} (2^{n} - 1) = 16 (2^{n} - 1)$
Since $n \geq 1$ , $2^{n} - 1 \neq = 0$ .
Cancel $(2^{n} - 1)$ from both sides:
$2^{a + 1} = 16$

Solving for $a$

Express $16$ as a power of $2$ : $16 = 2^{4}$
Equation becomes: $2^{a + 1} = 2^{4}$
Equate exponents: $a + 1 = 4$
Final Answer: $a = 3$

00:00 / 00:00

The Sigma Insight: Geometric Progression (G.P.)

Solution Diagram

Analyzing the Functional Equation

The given functional equation is

f (x + y) = f (x) f (y)

. This specific structure is the hallmark of exponential functions.

When you encounter this relation, your mind should immediately associate it with the form

f (x) = b^{x}

. This is because exponents satisfy the fundamental property:

b^{x + y} = b^{x} \cdot b^{y}

This is the unique family of functions that transforms addition in the input into multiplication in the output.

Finding the Heart of the Function

We are provided with the boundary condition

f (1) = 2

. If our function takes the form

f (x) = b^{x}

, then:

f (1) = b^{1} = 2

This reveals that the base

b

2

. Thus, the function is explicitly defined as:

f (x) = 2^{x}

The Summation Challenge

We are tasked with solving the summation:

k = 1 \sum n f (a + k) = 16 (2^{n} - 1)

Substituting

f (x) = 2^{x}

into the expression, we obtain:

k = 1 \sum n 2^{a + k} = 16 (2^{n} - 1)

To simplify, we use the laws of exponents to rewrite

2^{a + k}

2^{a} \cdot 2^{k}

. Since

2^{a}

is independent of the index

k

, we factor it out of the summation:

2^{a} k = 1 \sum n 2^{k} = 16 (2^{n} - 1)

The Geometric Progression

The remaining sum,

\sum_{k = 1}^{n} 2^{k} = 2^{1} + 2^{2} + \dots + 2^{n}

, is a classic Geometric Progression (G.P.). Here, the first term is

a^{'} = 2

and the common ratio is

r = 2

Using the sum formula for a G.P.,

S_{n} = \frac{a ^{'} ( r ^{n} - 1 )}{r - 1}

, we calculate:

S_{n} = \frac{2 ( 2 ^{n} - 1 )}{2 - 1} = 2 (2^{n} - 1)

The Final Convergence

Substituting the sum back into our equation, we get:

2^{a} \cdot 2 (2^{n} - 1) = 16 (2^{n} - 1)

Since

n

is a natural number, $(2^n - 1) eq 0$, allowing us to cancel this term from both sides. We are left with:

2^{a} \cdot 2 = 16

Applying the laws of exponents,

2^{a} \cdot 2^{1} = 2^{a + 1}

. Since

16 = 2^{4}

, the equation becomes:

2^{a + 1} = 2^{4}

Equating the exponents, we find

a + 1 = 4

. Therefore, the final result is: $a = 3$

Common Traps & FAQs

Wait, why does

f (x + y) = f (x) f (y)

automatically mean

f (x) = b^{x}

Great question! This is a Cauchy-type functional equation. If you test small values,

f (1 + 1) = f (1) f (1) = f (1)^{2}

f (1 + 1 + 1) = f (1)^{3}

, and so on. By induction,

f (n) = f (1)^{n}

. Since

f (1) = 2

, we get

f (n) = 2^{n}

. It is the unique bridge between additive inputs and multiplicative outputs.

Can I just expand the sum manually for

n = 1

and

n = 2

to find

a

You absolutely can! In JEE, if you are stuck, testing small values of

n

is a brilliant strategy. If

n = 1

, the sum is

f (a + 1) = 16 (2^{1} - 1) = 16

. Since

f (x) = 2^{x}

, we have

2^{a + 1} = 16

, so

a + 1 = 4

, which gives

a = 3

. It confirms the result instantly!

Why can we just cancel

(2^{n} - 1)

from both sides?

We can cancel it because

n

is a natural number (

n \geq 1

). For any

n \geq 1

2^{n}

is at least 2, so

2^{n} - 1

is at least 1. Since it is never zero, dividing both sides by this term is mathematically sound and perfectly safe.

Is there a risk that

f (x)

could be something other than

2^{x}

For natural numbers, the definition

f (x + y) = f (x) f (y)

with

f (1) = 2

uniquely defines the function as

2^{x}

. If the domain were real numbers, we would need to assume continuity to rule out pathological solutions, but for natural numbers, the exponential form is the only one that satisfies the condition.

Why did we pull

2^{a}

out of the summation?

The summation operator

\sum

acts only on the index

k

. Any term that does not contain

k

is treated as a constant relative to the summation. Since

2^{a + k} = 2^{a} \cdot 2^{k}

, the

2^{a}

is just a constant multiplier that can be moved to the front, simplifying the expression significantly.

Animated Solution for Mathematics - Sequence and Series: Find the natural number 'a' for which $\sum_{k = 1}^{n} f (a + k) = 16 (2^{n} - 1)$ , where the function 'f' satisfies the relation $f (x + y) = f (x) f (y)$ for all natural numbers $x, y$ and further $f (1) = 2$ .

Enter Numerical Value:

Visualized Solution

The Functional Equation

Deducing the Function Form

Using the Boundary Condition

Finding the Base $b$

The Summation Equation

Substituting $f (x)$ into the Sum

Splitting the Exponent

Factoring Out the Constant

Identifying the Geometric Progression

Applying the G.P. Sum Formula

Simplifying the G.P. Sum

Equating and Canceling Terms

Solving for $a$

The Sigma Insight: Geometric Progression (G.P.)

Analyzing the Functional Equation

Finding the Heart of the Function

The Summation Challenge

The Geometric Progression

The Final Convergence

Similar Questions

Let $\sum_{k = 1}^{10} f (a + k) = 16 (2^{10} - 1)$ , where the function $f$ satisfies $f (x + y) = f (x) f (y)$ for all natural numbers $x, y$ and $f (1) = 2$ . then the natural number 'a' is

Let $f (x)$ be a function such that $f (x + y) = f (x) \cdot f (y)$ for all $x, y \in N$ . If $f (1) = 3$ and $\sum_{k = 1}^{n} f (k) = 3279$ , then the value of $n$ is

If $f (x + y) = f (x) f (y)$ and $\sum_{x = 1}^{\infty} f (x) = 2$ , $x, y \in N$ where $N$ is the set of all natural numbers, then the value of $\frac{f ( 4 )}{f ( 2 )}$ is:

Let $f$ and $g$ be functions satisfying $f (x + y) = f (x) f (y), f (1) = 7$ and $g (x + y) = g (x y), g (1) = 1$ , for all $x, y \in N$ . If $\sum_{x=1}^{n} \left(\frac{f(x)}{g(x)} ight) = 19607 $, t h e n$ n$ is equal to:

If $a_{n} = \frac{3}{4} - (\frac{3}{4})^{2} + (\frac{3}{4})^{3} + \dots + (- 1)^{n - 1} (\frac{3}{4})^{n}$ and $b_{n} = 1 - a_{n}$ , then find the least natural number $n_{0}$ such that $b_{n} > a_{n}$ for all $n \geq n_{0}$ .

Let $a_{1} = 1$ and for $n \geq 1, a_{n + 1} = \frac{1}{2} a_{n} + \frac{n ^{2} - 2 n - 1}{n ^{2} ( n + 1 ) ^{2}}$ . Then $\sum_{n = 1}^{\infty} (a_{n} - \frac{2}{n ^{2}})$ is equal to ......... .

The greatest positive integer $k$ , for which $4 9^{k} + 1$ is a factor of the sum $4 9^{125} + 4 9^{124} + ... + 4 9^{2} + 49 + 1$ , is

The greatest positive integer $k$ , for which $4 9^{k} + 1$ is a factor of the sum $4 9^{125} + 4 9^{124} + \dots + 4 9^{2} + 49 + 1$ , is

Enter Numerical Value:

Visualized Solution

The Functional Equation

Deducing the Function Form

Using the Boundary Condition

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

The Summation Equation

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x) into the Sum

Splitting the Exponent

Factoring Out the Constant

Identifying the Geometric Progression

Applying the G.P. Sum Formula

Simplifying the G.P. Sum

Equating and Canceling Terms

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

The Sigma Insight: Geometric Progression (G.P.)

Analyzing the Functional Equation

Finding the Heart of the Function

The Summation Challenge

The Geometric Progression

The Final Convergence

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a1​=1 and for n≥1,an+1​=21​an​+n2(n+1)2n2−2n−1​. Then ∑n=1∞​(an​−n22​) is equal to ......... .

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The greatest positive integer k, for which 49k+1 is a factor of the sum 49125+49124+...+492+49+1, is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The greatest positive integer k, for which 49k+1 is a factor of the sum 49125+49124+⋯+492+49+1, is

Finding the Base $b$

Substituting $f (x)$ into the Sum

Solving for $a$

Let $\sum_{k = 1}^{10} f (a + k) = 16 (2^{10} - 1)$ , where the function $f$ satisfies $f (x + y) = f (x) f (y)$ for all natural numbers $x, y$ and $f (1) = 2$ . then the natural number 'a' is

Let $f (x)$ be a function such that $f (x + y) = f (x) \cdot f (y)$ for all $x, y \in N$ . If $f (1) = 3$ and $\sum_{k = 1}^{n} f (k) = 3279$ , then the value of $n$ is

If $f (x + y) = f (x) f (y)$ and $\sum_{x = 1}^{\infty} f (x) = 2$ , $x, y \in N$ where $N$ is the set of all natural numbers, then the value of $\frac{f ( 4 )}{f ( 2 )}$ is:

If $a_{n} = \frac{3}{4} - (\frac{3}{4})^{2} + (\frac{3}{4})^{3} + \dots + (- 1)^{n - 1} (\frac{3}{4})^{n}$ and $b_{n} = 1 - a_{n}$ , then find the least natural number $n_{0}$ such that $b_{n} > a_{n}$ for all $n \geq n_{0}$ .

Let $a_{1} = 1$ and for $n \geq 1, a_{n + 1} = \frac{1}{2} a_{n} + \frac{n ^{2} - 2 n - 1}{n ^{2} ( n + 1 ) ^{2}}$ . Then $\sum_{n = 1}^{\infty} (a_{n} - \frac{2}{n ^{2}})$ is equal to ......... .

The greatest positive integer $k$ , for which $4 9^{k} + 1$ is a factor of the sum $4 9^{125} + 4 9^{124} + ... + 4 9^{2} + 49 + 1$ , is

The greatest positive integer $k$ , for which $4 9^{k} + 1$ is a factor of the sum $4 9^{125} + 4 9^{124} + \dots + 4 9^{2} + 49 + 1$ , is