यदि \(Q_1=1\), \(Q_2=2\) और \(Q_n=Q_{n-1}+Q_{n-2}+n^2\) है, तो \(Q_5\) का मान क्या होगा?

If \(Q_1=1\), \(Q_2=2\), and \(Q_n=Q_{n-1}+Q_{n-2}+n^2\), what is the value of \(Q_5\)?

Author: Muft Shiksha Editorial Team Published:
Explanation opens after your attempt
Correct Answer

B. सड़सठ(67)

Step 1

Concept

\(Q_3=12\), \(Q_4=30\), and \(Q_5=67\). In exams, add \(n^2\) with both previous terms.

Step 2

Why this answer is correct

The correct answer is B. सड़सठ / (67). \(Q_3=12\), \(Q_4=30\), and \(Q_5=67\). In exams, add \(n^2\) with both previous terms.

Step 3

Exam Tip

\(Q_3=12\), \(Q_4=30\) और \(Q_5=67\) है। परीक्षा में दोनों पिछले पदों के साथ \(n^2\) जोड़ें।

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Mathematics Answer, Explanation and Revision Hints

यदि \(Q_1=1\), \(Q_2=2\) और \(Q_n=Q_{n-1}+Q_{n-2}+n^2\) है, तो \(Q_5\) का मान क्या होगा? / If \(Q_1=1\), \(Q_2=2\), and \(Q_n=Q_{n-1}+Q_{n-2}+n^2\), what is the value of \(Q_5\)?

Correct Answer: B. सड़सठ / (67). Explanation: \(Q_3=12\), \(Q_4=30\) और \(Q_5=67\) है। परीक्षा में दोनों पिछले पदों के साथ \(n^2\) जोड़ें। / \(Q_3=12\), \(Q_4=30\), and \(Q_5=67\). In exams, add \(n^2\) with both previous terms.

Which concept should I revise for this Mathematics MCQ?

\(Q_3=12\), \(Q_4=30\), and \(Q_5=67\). In exams, add \(n^2\) with both previous terms.

What exam hint can help solve this Mathematics question?

\(Q_3=12\), \(Q_4=30\) और \(Q_5=67\) है। परीक्षा में दोनों पिछले पदों के साथ \(n^2\) जोड़ें।