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Animated Solution for Mathematics - Sequence and Series: If $2^{10} + 2^{9} \cdot 3^{1} + 2^{8} \cdot 3^{2} + .... + 2 \cdot 3^{9} + 3^{10} = S - 2^{11}$ then $S$ is equal to :

Q: Wait, why is the number of terms $n=11$ and not $10$?

This is a classic trap! If you look at the powers of $3$, they start at $3^0$ (which is $1$) and go up to $3^{10}$. When counting terms from $0$ to $10$ inclusive, the count is $10 - 0 + 1 = 11$. Always double-check your index counting; it is the most common place to lose marks in competitive exams.

Q: What if I used the formula $\frac{a(1-r^n)}{1-r}$ instead?

That is perfectly valid! Since $r = \frac{3}{2} > 1$, using $\frac{a(r^n - 1)}{r - 1}$ is slightly more convenient to avoid dealing with negative signs in both the numerator and denominator, but both formulas are mathematically identical. You will arrive at the exact same result regardless of which version you choose.

Select Answer:

Visualized Solution

Analyze the Series Structure

Given series: $2^{10} + 2^{9} \cdot 3^{1} + 2^{8} \cdot 3^{2} + \dots + 2 \cdot 3^{9} + 3^{10}$
Observe the powers: $2$ decreases from $10$ to $0$ , while $3$ increases from $0$ to $10$ .
Let the sum be $L H S$ .

Identify the Geometric Progression

Check common ratio ( $r$ ):
$r = \frac{2 ^{9} \cdot 3 ^{1}}{2 ^{10}} = \frac{3}{2}$
$r = \frac{2 ^{8} \cdot 3 ^{2}}{2 ^{9} \cdot 3 ^{1}} = \frac{3}{2}$
Since the ratio is constant, it is a Geometric Progression.

Define G.P. Parameters

First term ( $a$ ) = $2^{10}$
Common ratio ( $r$ ) = $\frac{3}{2}$
Number of terms ( $n$ ): Powers of $3$ range from $0$ to $10$ , so $n = 10 - 0 + 1 = 11$ .

State the G.P. Sum Formula

Sum of G.P. formula: $S_{n} = \frac{a ( r ^{n} - 1 )}{r - 1}$

Substitute the Values

Substitute the values into the formula:
$L H S = \frac{2 ^{10} ( ( \frac{3}{2} ) ^{11} - 1 )}{\frac{3}{2} - 1}$

Simplify the Denominator

Denominator: $\frac{3}{2} - 1 = \frac{1}{2}$
Multiply by the reciprocal:
$L H S = 2^{10} \cdot 2 \cdot ((\frac{3}{2})^{11} - 1)$
$L H S = 2^{11} \cdot (\frac{3 ^{11}}{2 ^{11}} - 1)$

Distribute and Simplify

Distribute $2^{11}$ into the bracket:
$L H S = (2^{11} \cdot \frac{3 ^{11}}{2 ^{11}}) - (2^{11} \cdot 1)$
$L H S = 3^{11} - 2^{11}$

Compare with the Given Equation

Given equation: $L H S = S - 2^{11}$
Substitute our calculated $L H S$ :
$3^{11} - 2^{11} = S - 2^{11}$

Final Result

Cancel $- 2^{11}$ from both sides:
$S = 3^{11}$
Correct Option: (D)

00:00 / 00:00

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

Solution Diagram

Analyzing the Setup

The expression provided is

2^{10} + 2^{9} \cdot 3^{1} + 2^{8} \cdot 3^{2} + \dots + 2 \cdot 3^{9} + 3^{10}

. At first glance, it appears complex, but it follows a highly structured pattern.

Observe the powers of

2

: they start at

10

and decrease steadily to

0

. Simultaneously, the powers of

3

start at

0

and increase steadily to

10

Whenever you encounter terms where one variable's exponent decreases while another's increases, you are looking at a Geometric Progression.

Defining the Battlefield

To solve this, we must identify the three core components of the series: the first term (

a

), the common ratio (

r

), and the number of terms (

n

The first term is clearly

a = 2^{10}

. To find the common ratio, we divide the second term by the first:

r = \frac{2 ^{9} \cdot 3 ^{1}}{2 ^{10}} = \frac{3}{2}

Verifying with the next pair,

\frac{2 ^{8} \cdot 3 ^{2}}{2 ^{9} \cdot 3 ^{1}}

, we confirm the ratio is constant at

r = \frac{3}{2}

. Since the exponents range from

0

10

, the total number of terms is

n = 11

The Algebraic Dance

We utilize the standard sum formula for a Geometric Progression:

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

Substituting our identified values into the formula, we obtain:

L H S = \frac{2 ^{10} ( ( \frac{3}{2} ) ^{11} - 1 )}{\frac{3}{2} - 1}

The denominator simplifies to

\frac{3}{2} - 1 = \frac{1}{2}

. Dividing by

\frac{1}{2}

is equivalent to multiplying by

2

L H S = 2^{10} \cdot 2 \cdot ((\frac{3}{2})^{11} - 1) = 2^{11} (\frac{3 ^{11}}{2 ^{11}} - 1)

Distributing the

2^{11}

into the bracket, the terms cancel out:

L H S = 2^{11} \cdot \frac{3 ^{11}}{2 ^{11}} - 2^{11} = 3^{11} - 2^{11}

The Grand Finale

The problem states that this sum is equal to

S - 2^{11}

. We set up the final equation:

3^{11} - 2^{11} = S - 2^{11}

By adding

2^{11}

to both sides, the terms vanish, leaving us with the elegant result:

$S = 3^{11}$

Common Traps & FAQs

Wait, why is the number of terms

n = 11

and not

10

This is a classic trap! If you look at the powers of

3

, they start at

3^{0}

(which is

1

) and go up to

3^{10}

. When counting terms from

0

10

inclusive, the count is

10 - 0 + 1 = 11

. Always double-check your index counting; it is the most common place to lose marks in competitive exams.

Can I solve this without the G.P. formula?

Technically, yes, you could calculate each term individually, but that would be a massive waste of time and a recipe for calculation errors. The beauty of the G.P. formula is that it acts as a shortcut, collapsing a long, tedious addition into a single, elegant algebraic expression. In JEE Advanced, we value efficiency and pattern recognition over brute force.

What if I used the formula

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

instead?

That is perfectly valid! Since

r = \frac{3}{2} > 1

, using

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

is slightly more convenient to avoid dealing with negative signs in both the numerator and denominator, but both formulas are mathematically identical. You will arrive at the exact same result regardless of which version you choose.

Animated Solution for Mathematics - Sequence and Series: If $2^{10} + 2^{9} \cdot 3^{1} + 2^{8} \cdot 3^{2} + .... + 2 \cdot 3^{9} + 3^{10} = S - 2^{11}$ then $S$ is equal to :

Select Answer:

Visualized Solution

Analyze the Series Structure

Identify the Geometric Progression

Define G.P. Parameters

State the G.P. Sum Formula

Substitute the Values

Simplify the Denominator

Distribute and Simplify

Compare with the Given Equation

Final Result

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

Analyzing the Setup

Defining the Battlefield

The Algebraic Dance

The Grand Finale

Similar Questions

If $\frac{6}{3 ^{12}} + \frac{10}{3 ^{11}} + \frac{20}{3 ^{10}} + \frac{40}{3 ^{9}} + \dots + \frac{10240}{3} = 2^{n} \cdot m$ , where $m$ is odd, then $m \cdot n$ is equal to ________.

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 ......... .

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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; } If 210+29⋅31+28⋅32+....+2⋅39+310=S−211 then S is equal to :

Select Answer:

Visualized Solution

Analyze the Series Structure

Identify the Geometric Progression

Define G.P. Parameters

State the G.P. Sum Formula

Substitute the Values

Simplify the Denominator

Distribute and Simplify

Compare with the Given Equation

Final Result

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

Analyzing the Setup

Defining the Battlefield

The Algebraic Dance

The Grand Finale

Similar Questions

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Animated Solution for Mathematics - Sequence and Series: If $2^{10} + 2^{9} \cdot 3^{1} + 2^{8} \cdot 3^{2} + .... + 2 \cdot 3^{9} + 3^{10} = S - 2^{11}$ then $S$ is equal to :

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