The sum Σ(k=1 to 20) k/2^k is equal to- (1) 2 - 3/2^17 (2) 2 - 11/2^19 (3) 1 - 11/2^20 (4) 2 - 21/2^

Q: Wait, why do we multiply by $1/2$ and then subtract? What is the logic behind that?

Great question! The goal is to eliminate the 'arithmetic' part of the A.G.P. By multiplying by the common ratio $r = 1/2$ and shifting the terms, we align terms with the same denominator. When we subtract, the numerators become $2-1=1, 3-2=1$, and so on. This effectively turns the arithmetic-geometric series into a simple, pure geometric series that we already know how to sum!

Q: Why can't we just use the formula for the sum of an A.G.P. directly?

You absolutely can! The formula for the sum of an A.G.P. is derived exactly from this 'shift and subtract' method. However, in JEE Advanced, it is often safer and more insightful to derive it on the fly. It prevents memory errors and helps you understand the underlying mechanics, which is crucial if the problem has a slight twist.

Q: Why does the last term $- \frac{20}{2^{21}}$ appear by itself?

When you shift the series by one position to the right, the last term of the original series ($20/2^{20}$) moves to the position of the $21^{st}$ term in the shifted series. Since the original series only has $20$ terms, this $21^{st}$ term has no counterpart to subtract from, so it remains as a dangling term at the end of the subtraction. Don't let it intimidate you; it's just part of the process!

Question

The sum $\sum_{k=1}^{20} \frac{k}{2^k}$ is equal to-

Options:

(A) $2 - \frac{3}{2^{17}}$

(B) $2 - \frac{11}{2^{19}}$

(C) $1 - \frac{11}{2^{20}}$

(D) $2 - \frac{21}{2^{20}}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a mathematical puzzle that looks like a simple sum, but hides a beautiful, rhythmic structure. We are tasked with finding the sum $S = \sum_{k=1}^{20} \frac{k}{2^k}$.

At first glance, it is just a list of numbers: $\frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \dots + \frac{20}{2^{20}}$.

Look closer at the components. The numerators are $1, 2, 3, \dots, 20$, which form an Arithmetic Progression (A.P.). The denominators are $2, 4, 8, \dots, 2^{20}$, which form a Geometric Progression (G.P.). This is the hallmark of an **Arithmetic-Geometric Progression (A.G.P.)**.

### The Shift and Subtract Strategy

How do we tame this beast? The secret lies in the 'Shift and Subtract' method. We want to collapse the arithmetic part.

We know the common ratio of our G.P. is $r = \frac{1}{2}$. If we multiply our entire sum $S$ by this ratio, we get:

$$\frac{1}{2}S = \frac{1}{4} + \frac{2}{8} + \frac{3}{16} + \dots + \frac{20}{2^{21}}$$

Now, here is where the magic happens. We align the original sum $S$ and our new $\frac{1}{2}S$ such that terms with the same denominators are stacked.

When we subtract $\frac{1}{2}S$ from $S$, the numerators subtract perfectly: $2-1=1$, $3-2=1$, and so on. This leaves us with a beautiful, pure G.P.:

$$S - \frac{1}{2}S = \frac{1}{2} + \left(\frac{2}{4} - \frac{1}{4}ight) + \left(\frac{3}{8} - \frac{2}{8}ight) + \dots + \left(\frac{20}{2^{20}} - \frac{19}{2^{20}}ight) - \frac{20}{2^{21}}$$

### The Collapse

After the subtraction, the arithmetic part has vanished! We are left with:

$$\frac{1}{2}S = \left(\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots + \frac{1}{2^{20}}ight) - \frac{20}{2^{21}}$$

The terms inside the parentheses are a simple G.P. with first term $a = \frac{1}{2}$, common ratio $r = \frac{1}{2}$, and $n = 20$ terms. Using the sum formula $S_n = \frac{a(1-r^n)}{1-r}$, we get:

$$	ext{Sum} = \frac{\frac{1}{2}\left(1 - (\frac{1}{2})^{20}ight)}{1 - \frac{1}{2}} = 1 - \frac{1}{2^{20}}$$

### The Final Stretch

Now, we just substitute this back into our equation for $\frac{1}{2}S$:

$$\frac{1}{2}S = \left(1 - \frac{1}{2^{20}}ight) - \frac{20}{2^{21}}$$

Simplifying the last term, $\frac{20}{2^{21}} = \frac{10}{2^{20}}$. So, $\frac{1}{2}S = 1 - \frac{1}{2^{20}} - \frac{10}{2^{20}} = 1 - \frac{11}{2^{20}}$.

Finally, multiplying by $2$ to solve for $S$, we get the final result:

**$$S = 2 - \frac{11}{2^{19}}$$**

And there it is! The complexity dissolves into a simple, elegant result. Remember, in JEE Advanced, it is not just about the answer; it is about recognizing the pattern and trusting the process. You have mastered the A.G.P.!

Animated Solution for Mathematics - Sequence and Series: The sum $\sum_{k = 1}^{20} \frac{k}{2 ^{k}}$ is equal to-

Select Answer:

Visualized Solution

Identify the Series

The A.G.P. Strategy

Align and Subtract

Perform the Subtraction

Sum of the G.P.

Simplify the G.P. Sum

Combine and Simplify

Solve for S

The Sigma Insight: Arithmetic-Geometric Progression (A.G.P.)

Analyzing the Setup

The Shift and Subtract Strategy

The Collapse

The Final Stretch

Similar Questions

The sum of the infinite series $1 + \frac{2}{3} + \frac{7}{3 ^{2}} + \frac{12}{3 ^{3}} + \frac{17}{3 ^{4}} + \frac{22}{3 ^{5}} + \dots$ is equal to

If $(20)^{19} + 2 (21) (20)^{18} + 3 (21)^{2} (20)^{17} + \dots + 20 (21)^{19} = k (20)^{19}$ , then $k$ is equal to ____.

The sum $1 + 2 \cdot 3 + 3 \cdot 3^{2} + \dots + 10 \cdot 3^{9}$ is equal to

The sum to infinite term of the series $1 + \frac{2}{3} + \frac{6}{3 ^{2}} + \frac{10}{3 ^{3}} + \frac{14}{3 ^{4}} + \dots$ is

The sum of the infinite series $1 + \frac{5}{6} + \frac{12}{6 ^{2}} + \frac{22}{6 ^{3}} + \frac{35}{6 ^{4}} + \frac{51}{6 ^{5}} + \frac{70}{6 ^{6}} + \dots$ is equal to:

Let $S = 2 + \frac{6}{7} + \frac{12}{7 ^{2}} + \frac{20}{7 ^{3}} + \frac{30}{7 ^{4}} + \dots$ . then $4 S$ is equal to

Let $S = 109 + \frac{108}{5} + \frac{107}{5 ^{2}} + \dots + \frac{2}{5 ^{107}} + \frac{1}{5 ^{108}}$ . Then the value of $(16 S - (25)^{- 54})$ is equal to

If $(10)^{9} + 2 (11)^{1} (10)^{8} + 3 (11)^{2} (10)^{7} + \dots + 10 (11)^{9} = k (10)^{9}$ , then $k$ is equal to:

Let $a_{1}, a_{2}, a_{3}, \dots$ be an AP. If $\sum_{r = 1}^{\infty} \frac{a _{r}}{2 ^{r}} = 4$ , then $4 a_{2}$ is equal to ________.

If total number of runs scored in $n$ matches is $\frac{n + 1}{4} (2^{n + 1} - n - 2)$ where $n > 1$ , and the runs scored in the $k^{t h}$ match are given by $k \cdot 2^{n + 1 - k}$ , where $1 \leq k \leq n$ . Find $n$ .

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 sum ∑k=120​2kk​ is equal to-