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Question #40703

Difficulty: 8/10 | 🟢 Free

The Question

Let Sp(n)=∑k=1nkpS_p(n) = \sum_{k=1}^n k^p for p∈Np \in \mathbb{N}. Evaluate the following limit: lim⁡n→∞n2(17∣S5(n)n6−16−12n−12n2∣+19∣S6(n)n7−17−12n−34n2∣)\lim_{n \to \infty} n^2 \left( 17 \left| \frac{S_5(n)}{n^6} - \frac{1}{6} - \frac{1}{2n} - \frac{1}{2n^2} \right| + 19 \left| \frac{S_6(n)}{n^7} - \frac{1}{7} - \frac{1}{2n} - \frac{3}{4n^2} \right| \right)

Options

  • A.

    376\frac{37}{6}

  • B.

    914\frac{91}{4}

  • C.

    −376-\frac{37}{6}

  • D.

    674\frac{67}{4}

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

Sum of Powers of Natural Numbers|Indeterminate Forms|Domain Restrictions|Modulus Function Breakdown

Why This Question Is Tricky

Common Mistakes

This question tests the asymptotic expansion of the sum of powers of integers (Faulhaber's formula) and its interaction with the modulus function.

The primary trap lies in the domain restriction imposed by the modulus as n→∞n \to \infty.

Students often memorize only the first two terms of the power sum np+1p+1+np2\frac{n^{p+1}}{p+1} + \frac{n^p}{2} and either ignore the O(np−1)O(n^{p-1}) term or incorrectly assume it is a constant 16n2\frac{1}{6n^2} for all pp.

Furthermore, the expressions inside the moduli are deliberately constructed to be negative for large nn.

If a student blindly drops the modulus without analyzing the sign of the dominant 1/n21/n^2 term, they will obtain the negative of the correct answer.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterSequences and Series - Class 11
ConceptSum of Powers of Natural Numbers
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree