JEE Main 2020 (7 January Shift 1)

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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 $4 9^{125} + 4 9^{124} + \dots + 4 9^{2} + 49 + 1$ , is

Q: Wait, why is the number of terms $126$ and not $125$? The power goes up to $125$!

This is the most common trap! The series starts at $49^0$ (which is $1$) and ends at $49^{125}$. Counting from $0$ to $125$ gives $125 - 0 + 1 = 126$ terms. Always check your boundaries, especially when the series starts with $1$!

Q: How do I know $k=63$ is the greatest integer? Could it be higher?

Great question. Our factorization $(49^{63} + 1)(49^{63} - 1)$ is derived from the difference of squares. If we tried to go higher, we would lose the structure that allows the Binomial expansion to cancel out the constant term. $63$ is the limit of this specific algebraic path.

Select Answer:

Visualized Solution

Identify the Series $S$

Given series: $S = 1 + 49 + 4 9^{2} + \dots + 4 9^{125}$
Notice the pattern: each term is multiplied by $49$ .
This forms a Geometric Progression (G.P.).

Parameters of the G.P.

First term, $a = 1$
Common ratio, $r = 49$
Powers of $49$ range from $0$ to $125$ .
Total number of terms, $n = 126$ .

Apply G.P. Sum Formula

Sum of G.P.: $S = \frac{a ( r ^{n} - 1 )}{r - 1}$
Substitute $a = 1$ , $r = 49$ , $n = 126$ :
$S = \frac{1 \cdot ( 4 9 ^{126} - 1 )}{49 - 1}$

Simplify the Expression

Denominator: $49 - 1 = 48$
Simplified Sum: $S = \frac{4 9 ^{126} - 1}{48}$

Factorize using Difference of Squares

We need a factor of the form $4 9^{k} + 1$ .
Recall the identity: $x^{2} - y^{2} = (x + y) (x - y)$
Express $4 9^{126}$ as a square: $4 9^{126} = (4 9^{63})^{2}$

Apply Factorization

Numerator becomes: $(4 9^{63})^{2} - 1^{2}$
Apply identity: $(4 9^{63} + 1) (4 9^{63} - 1)$
Substitute back into $S$ :
$S = \frac{( 4 9 ^{63} + 1 ) ( 4 9 ^{63} - 1 )}{48}$

Analyze Divisibility

For $4 9^{63} + 1$ to be a factor of integer $S$ , the remaining part must be an integer.
Remaining part: $\frac{4 9 ^{63} - 1}{48}$
We must prove that $48$ perfectly divides $4 9^{63} - 1$ .

Binomial Expansion for Divisibility

Rewrite $49$ as $(1 + 48)$
Numerator: $(1 + 48)^{63} - 1$
Expand using Binomial Theorem:
$[1 + (1 63) 48 + (2 63) 4 8^{2} + \dots + 4 8^{63}] - 1$

Simplify the Expansion

The $+ 1$ and $- 1$ cancel out.
Remaining terms: $(1 63) 48 + (2 63) 4 8^{2} + \dots + 4 8^{63}$
Factor out $48$ :
$48 \times [(1 63) + (2 63) 48 + \dots + 4 8^{62}]$

Final Conclusion

Since $4 9^{63} - 1 = 48 \times I$ , then $\frac{4 9 ^{63} - 1}{48} = I$ (an integer).
Therefore, $S = (4 9^{63} + 1) \times I$
Comparing with $4 9^{k} + 1$ , the greatest integer $k = 63$ .

00:00 / 00:00

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

Analyzing the Setup

The given expression is a sum of a geometric series:

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

Each term is obtained by multiplying the previous term by a common ratio

r = 49

. The first term is

a = 1

Since the powers of

49

range from

0

125

, the total number of terms is

n = 126

. Using the sum formula for a geometric progression,

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

, we obtain:

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

The Algebraic Surgeon

We aim to identify a factor of the form

4 9^{k} + 1

. To achieve this, we manipulate the numerator using the difference of squares identity,

x^{2} - y^{2} = (x + y) (x - y)

Recognizing that

126 = 2 \times 63

, we rewrite the numerator as:

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

Applying the identity, we factor the expression:

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

Thus, the sum

S

can be expressed as:

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

The Final Proof of Divisibility

To confirm that

4 9^{63} + 1

is a factor of

S

, we must verify that the term

I = \frac{4 9 ^{63} - 1}{48}

is an integer. We utilize the Binomial Theorem by expressing

49

(1 + 48)

Consider the expansion of

(1 + 48)^{63}

(1 + 48)^{63} = 1 + (1 63) (48) + (2 63) (48)^{2} + \dots + (48)^{63}

Subtracting

1

from both sides, we get:

(1 + 48)^{63} - 1 = (1 63) (48) + (2 63) (48)^{2} + \dots + (48)^{63}

Every term on the right-hand side contains at least one factor of

48

. Therefore, the expression is divisible by

48

, making

I

an integer.

Comparing our result

S = (4 9^{63} + 1) \times I

to the required form

4 9^{k} + 1

, we conclude that:

$k = 63$

Common Traps & FAQs

Wait, why is the number of terms

126

and not

125

? The power goes up to

125

This is the most common trap! The series starts at

4 9^{0}

(which is

1

) and ends at

4 9^{125}

. Counting from

0

125

gives

125 - 0 + 1 = 126

terms. Always check your boundaries, especially when the series starts with

1

Why do we need to prove that

\frac{4 9 ^{63} - 1}{48}

is an integer? Can't we just assume it?

In JEE Advanced, rigor is everything. A factor must be an integer. If the remaining part were a fraction, then

4 9^{63} + 1

wouldn't be a factor of the integer

S

. We prove it's an integer to ensure our factorization is valid and mathematically sound.

How do I know

k = 63

is the greatest integer? Could it be higher?

Great question. Our factorization

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

is derived from the difference of squares. If we tried to go higher, we would lose the structure that allows the Binomial expansion to cancel out the constant term.

63

is the limit of this specific algebraic path.

Animated Solution for Mathematics - Sequence and Series: The greatest positive integer $k$ , for which $4 9^{k} + 1$ is a factor of the sum $4 9^{125} + 4 9^{124} + \dots + 4 9^{2} + 49 + 1$ , is

Select Answer:

Visualized Solution

Identify the Series $S$

Parameters of the G.P.

Apply G.P. Sum Formula

Simplify the Expression

Factorize using Difference of Squares

Apply Factorization

Analyze Divisibility

Binomial Expansion for Divisibility

Simplify the Expansion

Final Conclusion

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

Analyzing the Setup

The Algebraic Surgeon

The Final Proof of Divisibility

Similar Questions

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

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 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 :

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

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 $a_{1}, a_{2}, a_{3}, ...$ 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:

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

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

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

Parameters of the G.P.

Apply G.P. Sum Formula

Simplify the Expression

Factorize using Difference of Squares

Apply Factorization

Analyze Divisibility

Binomial Expansion for Divisibility

Simplify the Expansion

Final Conclusion

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

Analyzing the Setup

The Algebraic Surgeon

The Final Proof of Divisibility

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 :

.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:

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

Animated Solution for Mathematics - Sequence and Series: The greatest positive integer $k$ , for which $4 9^{k} + 1$ is a factor of the sum $4 9^{125} + 4 9^{124} + \dots + 4 9^{2} + 49 + 1$ , is

Identify the Series $S$

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

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: