MCQMediumJEE 2023Sum of Series

JEE Mathematics 2023 Question with Solution

Let SK=1+2++KKS_K = \frac{1 + 2 + \dots + K}{K} and j=1nSj2\sum_{j=1}^{n} S_j^2 where A,B,C,DNA, B, C, D \in \mathbb{N} and AA has the least value. Then:

  • A

    A+BA + B is divisible by DD

  • B

    A+BA + B is divisible by 55

  • C

    A+C+DA + C + D is not divisible by BB

  • D

    A+B+DA + B + D is divisible by 55

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: SK=1+2++KKS_K = \frac{1 + 2 + \dots + K}{K}.

Find: Which statement about A,B,C,DA, B, C, D is correct.

From the solution,

SK=K+12S_K = \frac{K+1}{2}

and hence

SK2=(K+1)24=K2+2K+14.S_K^2 = \frac{(K+1)^2}{4} = \frac{K^2 + 2K + 1}{4}.

The extracted solution then states that on simplifying the summation and comparing the required terms, we obtain

A=24,B=2,C=9,D=13.A = 24, \quad B = 2, \quad C = 9, \quad D = 13.

Now check the options using these values:

A+B=24+2=26A + B = 24 + 2 = 26

and

D=13.D = 13.

Therefore,

26 is divisible by 13.26 \text{ is divisible by } 13.

So the correct statement is A+BA + B is divisible by DD. Therefore, the correct option is A.

Using the extracted values directly

Given: the solution provides the final values A=24,B=2,C=9,D=13A = 24, B = 2, C = 9, D = 13.

Find: Which option matches these values.

Evaluate each relevant expression:

A+B=26,A+C+D=24+9+13=46,A+B+D=39.A + B = 26, \quad A + C + D = 24 + 9 + 13 = 46, \quad A + B + D = 39.

Now test the divisibility claims:

  • 2626 is divisible by 1313.
  • 2626 is not divisible by 55.
  • 4646 is divisible by 22, so the statement "not divisible by BB" is false.
  • 3939 is not divisible by 55.

Hence only the first statement is true. Therefore, the correct option is A.

Common mistakes

  • Using the option letter shown in the solution without checking the numbered options. The page says option B, but the provided correct-answer field clearly maps statement (1) to label A. Always map source numbering (1),(2),(3),(4)(1),(2),(3),(4) to labels A,B,C,DA,B,C,D carefully.

  • Confusing SKS_K with the sum 1+2++K1+2+\dots+K itself. Here SKS_K is that sum divided by KK, so first compute 1+2++K=K(K+1)21+2+\dots+K = \frac{K(K+1)}{2} and then divide by KK to get K+12\frac{K+1}{2}.

  • Checking only one numerical relation incompletely. After obtaining A,B,C,DA, B, C, D, verify divisibility exactly rather than by approximation. For example, test whether 26÷1326 \div 13 is an integer, not whether the numbers merely look related.

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