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NVAMediumJEE 2023Sum of Series

JEE Mathematics 2023 Question with Solution

The sum 12−2⋅32+3.52−4.72+5.92−⋯+15.2921^2 - 2 \cdot 3^2 + 3.5^2 - 4.7^2 + 5.9^2 - \dots + 15.29^2, is:

Answer

Correct answer:6952

Step-by-step solution

Standard Method

Given:

S=12−2⋅32+3.52−4.72+5.92−⋯+15.292S = 1^2 - 2 \cdot 3^2 + 3.5^2 - 4.7^2 + 5.9^2 - \dots + 15.29^2

Find: The value of SS.

Separate the odd placed and even placed terms:

S=(1.12+3.52+⋯+15.292)−(2.32+4.72+⋯+14.272)S = (1.1^2 + 3.5^2 + \dots + 15.29^2) - (2.3^2 + 4.7^2 + \dots + 14.27^2)

Write these terms in summation form:

S=∑n=18(2n−1)(4n−3)2−∑n=17(2n)(4n−1)2S = \sum_{n=1}^{8} (2n-1)(4n-3)^2 - \sum_{n=1}^{7} (2n)(4n-1)^2

Applying the summation formula, we get

S=29856−22904=6952S = 29856 - 22904 = 6952

Therefore, the value of the sum is 69526952.

Separated-Series Form

Given:

S=12−2⋅32+3.52−4.72+5.92−⋯+15.292S = 1^2 - 2 \cdot 3^2 + 3.5^2 - 4.7^2 + 5.9^2 - \dots + 15.29^2

Find: The numerical value of the alternating sum.

The pattern is separated into positive odd-placed terms and negative even-placed terms as shown in the solution.

So,

S=(12+3.52+⋯+15.292)−(22+4.72+⋯+14.272)S = (1^2 + 3.5^2 + \dots + 15.29^2) - (2^2 + 4.7^2 + \dots + 14.27^2)

Using the indexed form given:

S=∑n=18(2n−1)2⋅(4n−3)2−∑n=17(2n)(4n−1)2S = \sum_{n=1}^{8} (2n-1)^2 \cdot (4n-3)^2 - \sum_{n=1}^{7} (2n)(4n-1)^2

And then,

S=∑n=18(2n−1)(4n−3)2−∑n=17(2n)(4n−1)2=29856−22904S = \sum_{n=1}^{8} (2n-1)(4n-3)^2 - \sum_{n=1}^{7} (2n)(4n-1)^2 = 29856 - 22904

Hence,

S=6952S = 6952

So, the correct answer is 69526952.

Common mistakes

  • Treating the expression as a simple alternating sum of squares only is incorrect, because coefficients such as 2,3,4,5,…2, 3, 4, 5, \dots are part of the terms. First identify the actual pattern in each term before applying summation formulas.

  • Mixing the odd-placed and even-placed terms leads to wrong limits of summation. The positive group has 88 terms up to 15.29215.29^2, while the negative group has 77 terms up to 14.27214.27^2.

  • Using incorrect general terms for the two sequences is a conceptual error. The solution rewrites the positive terms with (2n−1)(4n−3)2(2n-1)(4n-3)^2 and the negative terms with (2n)(4n−1)2(2n)(4n-1)^2, so the indexing must match the observed pattern.

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