Visual SolutionPYQ 2023 · Jan Shift 1Standard

Question

The sum of the series

1 + 2 \cdot 2 + 3 \cdot 2^2 + 4 \cdot 2^3 + \ldots + 100 \cdot 2^{99}

is:

(A)

99 \cdot 2^{100} + 1

(B)

100 \cdot 2^{100} + 1

(C)

99 \cdot 2^{100} - 1

(D)

100 \cdot 2^{99} + 1

Solution Path

AGP method: multiply by

r = 2

, subtract to telescope, evaluate remaining GP sum

\to 99 \cdot 2^{100} + 1

01Question Setup

1/4

Find the sum

S = 1 + 2 \cdot 2 + 3 \cdot 2^2 + 4 \cdot 2^3 + \ldots + 100 \cdot 2^{99}

. This is an Arithmetico-Geometric Progression (AGP) with AP part

k

and GP part

2^{k-1}

S = \sum_{k=1}^{100} k \cdot 2^{k-1}

, an AGP with

r = 2

02Multiply-Shift-SubtractKEY INSIGHT

2/4

Multiply

S

r = 2

to get

2S

, then align terms and subtract

S

from

2S

. All inner terms telescope, leaving only the boundary terms.

2S - S = 100 \cdot 2^{100} - (2 + 2^2 + \ldots + 2^{99}) - 1

03Evaluate the GP

3/4

The remaining GP sum

1 + 2 + 2^2 + \ldots + 2^{99} = 2^{100} - 1

by the standard formula

S_n = a(r^n - 1)/(r - 1)

S = 100 \cdot 2^{100} - (2^{100} - 1) = 99 \cdot 2^{100} + 1

04Final Answer

4/4

The AGP method reduces an infinite-looking sum to a clean closed form in just three steps: multiply, shift, subtract.

S = 99 \cdot 2^{100} + 1

, Answer: (A)

Concepts from this question2 concepts unlocked

★

EASY

Sum of n terms of a GP with first term a and common ratio r

S_n = a \cdot \frac{r^n - 1}{r - 1} \quad (r \neq 1)

Foundation for nearly every GP problem in JEE. Must be instant recall.

Sum problemsRatio conditionsInfinite GP

STANDARD

When dividing two GP sums with the same base, a and r^n terms cancel, leaving a simple expression in r

\frac{S}{S_{\text{odd}}} = \frac{r^2 - 1}{r - 1} = r + 1

The 'messy' parts (a, r^n) always cancel. Recognizing this saves 3-4 minutes of algebra.

Sum ratio problemsCondition-based GPInteger ratio constraints

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