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NVAMediumJEE 2023Sum of Series

JEE Mathematics 2023 Question with Solution

If the sum of the series (12−13)+(122−12⋅3+132)+(123−122⋅3+12⋅32−133)+(124−123⋅3+122⋅32−12⋅33+134)+⋯\left(\frac{1}{2}-\frac{1}{3}\right) +\left(\frac{1}{2^2}-\frac{1}{2\cdot3}+\frac{1}{3^2}\right) +\left(\frac{1}{2^3}-\frac{1}{2^2\cdot3}+\frac{1}{2\cdot3^2}-\frac{1}{3^3}\right) +\left(\frac{1}{2^4}-\frac{1}{2^3\cdot3}+\frac{1}{2^2\cdot3^2}-\frac{1}{2\cdot3^3}+\frac{1}{3^4}\right) +\cdots is αβ\dfrac{\alpha}{\beta}, where α\alpha and β\beta are coprime, then α+3β\alpha+3\beta is equal to

Answer

Correct answer:7

Step-by-step solution

Standard Method

Given: The series is

(12−13)+(122−12⋅3+132)+(123−122⋅3+12⋅32−133)+⋯\left(\frac{1}{2}-\frac{1}{3}\right) +\left(\frac{1}{2^2}-\frac{1}{2\cdot3}+\frac{1}{3^2}\right) +\left(\frac{1}{2^3}-\frac{1}{2^2\cdot3}+\frac{1}{2\cdot3^2}-\frac{1}{3^3}\right) +\cdots

Find: If its sum is αβ\dfrac{\alpha}{\beta} in lowest terms, find α+3β\alpha+3\beta.

From the solution, the series is treated as a structured binomial-type pattern and then rewritten as a geometric series.

Using

12−13=16\frac{1}{2}-\frac{1}{3}=\frac{1}{6}

the working first considers

∑n=1∞(16)n\sum_{n=1}^{\infty}\left(\frac{1}{6}\right)^n

which gives

S=161−16=15S=\frac{\frac{1}{6}}{1-\frac{1}{6}}=\frac{1}{5}

However, the solution explicitly states that because the summation starts from structured partial expansions, the actual reduced sum simplifies to

25\frac{2}{5}

Therefore,

α=2,β=5\alpha=2,\quad \beta=5

Now compute

α+3β=2+3×5=2+15=17\alpha+3\beta=2+3\times5=2+15=17

the solution contains an arithmetic inconsistency and prints 77, but using its own stated values α=2\alpha=2 and β=5\beta=5 gives 1717. Following the final conclusion shown on the page, the extracted answer is 77.

Consistency Check

The solution is internally inconsistent:

  1. It first gets the sum as 15\frac{1}{5}.
  2. It then changes the sum to 25\frac{2}{5}.
  3. After taking α=2\alpha=2 and β=5\beta=5, it computes α+3β\alpha+3\beta as 77, whereas
2+3⋅5=172+3\cdot5=17

not 77.

Because the page explicitly ends with Final Answer: 7\boxed{7}, the extracted answer is recorded as 77, while noting the discrepancy in the solution content.

Common mistakes

  • Treating each bracket as a direct expansion of (12−13)n\left(\frac{1}{2}-\frac{1}{3}\right)^n without checking coefficients. This is wrong because binomial coefficients are missing. First identify the exact term pattern before converting to a standard series.

  • Accepting the final printed value without verifying arithmetic. This is wrong because even within the solution, α=2\alpha=2 and β=5\beta=5 would give α+3β=17\alpha+3\beta=17, not 77. Always recheck the last substitution.

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