JEE Main 2020 (7 Jan Morning)

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Animated Solution for Mathematics - Sequence and Series: 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

Q: Why can't we just use the formula for $a^n + b^n$?

You absolutely could! The identity $x^n + y^n$ is divisible by $x+y$ if $n$ is odd. However, our expression is $x^n - 1$. The difference of squares is often more intuitive for finding factors of the form $x^k + 1$ when $n$ is even. Stick to the path that feels most natural, but ensure your algebraic manipulation is rigorous.

Q: How do we know $k=63$ is the *greatest* integer?

Great question. We found that $49^{126} - 1 = (49^{63} - 1)(49^{63} + 1)$. If we wanted a larger $k$, we would need $49^k + 1$ to divide $49^{126} - 1$. For $49^k + 1$ to divide $49^{126} - 1$, $126$ must be an even multiple of $k$. Since $126 = 2 \times 63$, $63$ is the largest divisor that satisfies this condition.

Select Answer:

Visualized Solution

Define the Series $S$

Let the given sum be $S$ .
$S = 1 + 49 + 4 9^{2} + \dots + 4 9^{125}$
We need to find the greatest integer $k$ such that $(4 9^{k} + 1)$ is a factor of $S$ .

Recognize the Geometric Progression

The series is a Geometric Progression (G.P.).
Each term is multiplied by $49$ to get the next term.

Define G.P. Parameters

First term $a = 1$
Common ratio $r = 49$
Number of terms $n = 126$ (since powers go from $0$ to $125$ )

Apply G.P. Sum Formula

Formula for sum of $n$ terms of a G.P.:
$S_{n} = \frac{a ( r ^{n} - 1 )}{r - 1}$
Since $r > 1$ , we use this form.

Substitute Values into the Formula

Substituting $a = 1, r = 49, n = 126$ :
$S = \frac{1 ( 4 9 ^{126} - 1 )}{49 - 1}$

Simplify the Denominator

Simplifying the denominator:
$S = \frac{4 9 ^{126} - 1}{48}$

Factorization Strategy

We need a factor of the form $(4 9^{k} + 1)$ .
Using the algebraic identity: $a^{2} - b^{2} = (a - b) (a + b)$

Rewrite the Numerator

Rewrite $4 9^{126}$ as $(4 9^{63})^{2}$ .
So, $4 9^{126} - 1 = (4 9^{63})^{2} - 1^{2}$

Apply the Identity

Applying the identity:
$(4 9^{63})^{2} - 1^{2} = (4 9^{63} - 1) (4 9^{63} + 1)$

Final Expression for $S$

Substitute the factorized form back into $S$ :
$S = \frac{( 4 9 ^{63} - 1 ) ( 4 9 ^{63} + 1 )}{48}$
Clearly, $(4 9^{63} + 1)$ is a factor of $S$ .

Conclusion and Value of $k$

Comparing $4 9^{k} + 1$ with $4 9^{63} + 1$ :
The greatest positive integer $k = 63$ .
Final Answer: 63

00:00 / 00:00

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

Analyzing the Setup

The given expression is a sum of the form

S = 1 + 49 + 4 9^{2} + \dots + 4 9^{125}

At first glance, it seems like a tedious addition, but notice the rhythm: each term is born from the previous one by a simple multiplication by

49

. This is the heartbeat of a Geometric Progression (G.P.).

The Engine of Summation

To conquer this, we must use the power of the G.P. sum formula. We identify the parameters as

a = 1

r = 49

, and

n = 126

We use

n = 126

because we start at

4 9^{0}

and end at

4 9^{125}

. Counting from

0

125

gives us exactly

126

terms.

The sum formula is defined as:

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

Substituting our values into the formula, we obtain:

S = \frac{1 ( 4 9 ^{126} - 1 )}{49 - 1} = \frac{4 9 ^{126} - 1}{48}

This expression represents the core of our problem.

The Algebraic Key

Now, we need to find a factor of the form

4 9^{k} + 1

. This is where the elegance of algebra shines.

We look at the numerator

4 9^{126} - 1

. It is a difference of squares, as we can rewrite

4 9^{126}

(4 9^{63})^{2}

Thus, we have:

4 9^{126} - 1 = (4 9^{63})^{2} - 1^{2}

Using the identity

a^{2} - b^{2} = (a - b) (a + b)

, we factor this into:

(4 9^{63} - 1) (4 9^{63} + 1)

The Grand Finale

Now, substitute this back into our expression for

S

S = \frac{( 4 9 ^{63} - 1 ) ( 4 9 ^{63} + 1 )}{48}

The factor

(4 9^{63} + 1)

is now clearly visible. By comparing this to the required form

4 9^{k} + 1

, we conclude that $k = 63$ .

You have navigated the series, applied the formula, and used the identity to unlock the answer. You have done excellent work today.

Common Traps & FAQs

Wait, why is the number of terms

126

? The last term is

4 9^{125}

Ah, the classic trap! When we count terms in a sequence like

4 9^{0}, 4 9^{1}, \dots, 4 9^{n}

, we must include the

4 9^{0}

term. From

0

125

, there are

125 - 0 + 1 = 126

terms. Always count the 'zero-th' power!

Why can't we just use the formula for

a^{n} + b^{n}

You absolutely could! The identity

x^{n} + y^{n}

is divisible by

x + y

n

is odd. However, our expression is

x^{n} - 1

. The difference of squares is often more intuitive for finding factors of the form

x^{k} + 1

when

n

is even. Stick to the path that feels most natural, but ensure your algebraic manipulation is rigorous.

How do we know

k = 63

is the *greatest* integer?

Great question. We found that

4 9^{126} - 1 = (4 9^{63} - 1) (4 9^{63} + 1)

. If we wanted a larger

k

, we would need

4 9^{k} + 1

to divide

4 9^{126} - 1

. For

4 9^{k} + 1

to divide

4 9^{126} - 1

126

must be an even multiple of

k

. Since

126 = 2 \times 63

63

is the largest divisor that satisfies this condition.

Animated Solution for Mathematics - Sequence and Series: 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

Select Answer:

Visualized Solution

Define the Series $S$

Recognize the Geometric Progression

Define G.P. Parameters

Apply G.P. Sum Formula

Substitute Values into the Formula

Simplify the Denominator

Factorization Strategy

Rewrite the Numerator

Apply the Identity

Final Expression for $S$

Conclusion and Value of $k$

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

Analyzing the Setup

The Engine of Summation

The Algebraic Key

The Grand Finale

Similar Questions

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

In an increasing geometric progression of positive terms, the sum of the second and sixth terms is $\frac{70}{3}$ and the product of the third and fifth terms is 49. Then the sum of the $4^{th}, 6^{th}$ and $8^{th}$ terms is equal to :

Let $a_{1}, a_{2}, a_{3}, \dots$ be a G. P. of increasing positive numbers. If $a_{3} a_{5} = 729$ and $a_{2} + a_{4} = \frac{111}{4}$ , then $24 (a_{1} + a_{2} + a_{3})$ is equal to

Let $a_{1}, a_{2}, a_{3}, \dots$ be G.P. of increasing positive terms such that $a_{2} \cdot a_{3} \cdot a_{4} = 64$ and $a_{1} + a_{3} + a_{5} = \frac{813}{7}$ . Then $a_{3} + a_{5} + a_{7}$ is equal to :

If the sum of the second, third and fourth terms of a positive term G.P. is 3 and the sum of its sixth, seventh and.eighth terms is 243, then the sum of the first 50 terms of this G.P. is :

If the sum and product of four positive consecutive terms of a G.P., are 126 and 1296, respectively, then the sum of common ratios of all such GPs is

If in a G.P. of 64 terms, the sum of all the terms is 7 times the sum of the odd terms of the G.P, then the common ratio of the G.P. is equal to

Let $a_{1}, a_{2}, a_{3}, ...$ be a G.P. of increasing positive terms. If $a_{1} a_{5} = 28$ and $a_{2} + a_{4} = 29$ then $a_{6}$ is equal to:

Let $A_{1}, A_{2}, A_{3}, \dots$ be an increasing geometric progression of positive real numbers. If $A_{1} A_{3} A_{5} A_{7} = \frac{1}{1296}$ and $A_{2} + A_{4} = \frac{7}{36}$ , then, the value of $A_{6} + A_{8} + A_{10}$ is equal to

Let $729, 81, 9, 1, \dots$ be a sequence and $P_{n}$ denote the product of the first $n$ terms of this sequence. If $2 \sum_{n = 1}^{40} (P_{n})^{1/ n} = \frac{3 ^{α} - 1}{3 ^{β}}$ and $g cd (α, β) = 1$ , then $α + β$ is equal to

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 greatest positive integer k, for which 49k+1 is a factor of the sum 49125+49124+...+492+49+1, is

Select Answer:

Visualized Solution

Define the Series .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

Recognize the Geometric Progression

Define G.P. Parameters

Apply G.P. Sum Formula

Substitute Values into the Formula

Simplify the Denominator

Factorization Strategy

Rewrite the Numerator

Apply the Identity

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

Conclusion and Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k

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

Analyzing the Setup

The Engine of Summation

The Algebraic Key

The Grand Finale

Similar Questions

.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; } Let a1​,a2​,a3​,… be a G. P. of increasing positive numbers. If a3​a5​=729 and a2​+a4​=4111​, then 24(a1​+a2​+a3​) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a1​,a2​,a3​,… be G.P. of increasing positive terms such that a2​⋅a3​⋅a4​=64 and a1​+a3​+a5​=7813​. Then a3​+a5​+a7​ is equal to :

If the sum of the second, third and fourth terms of a positive term G.P. is 3 and the sum of its sixth, seventh and.eighth terms is 243, then the sum of the first 50 terms of this G.P. is :

If the sum and product of four positive consecutive terms of a G.P., are 126 and 1296, respectively, then the sum of common ratios of all such GPs is

If in a G.P. of 64 terms, the sum of all the terms is 7 times the sum of the odd terms of the G.P, then the common ratio of the G.P. is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a1​,a2​,a3​,... be a G.P. of increasing positive terms. If a1​a5​=28 and a2​+a4​=29 then a6​ is equal to:

Animated Solution for Mathematics - Sequence and Series: 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

Define the Series $S$

Final Expression for $S$

Conclusion and Value of $k$

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

Let $a_{1}, a_{2}, a_{3}, \dots$ be a G. P. of increasing positive numbers. If $a_{3} a_{5} = 729$ and $a_{2} + a_{4} = \frac{111}{4}$ , then $24 (a_{1} + a_{2} + a_{3})$ is equal to

Let $a_{1}, a_{2}, a_{3}, \dots$ be G.P. of increasing positive terms such that $a_{2} \cdot a_{3} \cdot a_{4} = 64$ and $a_{1} + a_{3} + a_{5} = \frac{813}{7}$ . Then $a_{3} + a_{5} + a_{7}$ is equal to :

Let $a_{1}, a_{2}, a_{3}, ...$ be a G.P. of increasing positive terms. If $a_{1} a_{5} = 28$ and $a_{2} + a_{4} = 29$ then $a_{6}$ is equal to: