यदि \(x_1=4\) और \(x_n=2x_{n-1}+n^2-n\) है, तो \(x_4\) का मान क्या होगा?

If \(x_1=4\) and \(x_n=2x_{n-1}+n^2-n\), what is the value of \(x_4\)?

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

A. चौंसठ(64)

Step 1

Concept

\(x_2=10\), \(x_3=26\), and \(x_4=64\). In exams, simplify \(n^2-n\) before adding it.

Step 2

Why this answer is correct

The correct answer is A. चौंसठ / (64). \(x_2=10\), \(x_3=26\), and \(x_4=64\). In exams, simplify \(n^2-n\) before adding it.

Step 3

Exam Tip

\(x_2=10\), \(x_3=26\) और \(x_4=64\) है। परीक्षा में \(n^2-n\) को सरल करके जोड़ें।

Question me issue ya doubt hai?

Answer, explanation, typing mistake ya suggestion directly hamari team ko bhejein. 📱Helpline (Call / WhatsApp): +91 7272824365

Related Mathematics Questions

FAQs

Mathematics Answer, Explanation and Revision Hints

यदि \(x_1=4\) और \(x_n=2x_{n-1}+n^2-n\) है, तो \(x_4\) का मान क्या होगा? / If \(x_1=4\) and \(x_n=2x_{n-1}+n^2-n\), what is the value of \(x_4\)?

Correct Answer: A. चौंसठ / (64). Explanation: \(x_2=10\), \(x_3=26\) और \(x_4=64\) है। परीक्षा में \(n^2-n\) को सरल करके जोड़ें। / \(x_2=10\), \(x_3=26\), and \(x_4=64\). In exams, simplify \(n^2-n\) before adding it.

Which concept should I revise for this Mathematics MCQ?

\(x_2=10\), \(x_3=26\), and \(x_4=64\). In exams, simplify \(n^2-n\) before adding it.

What exam hint can help solve this Mathematics question?

\(x_2=10\), \(x_3=26\) और \(x_4=64\) है। परीक्षा में \(n^2-n\) को सरल करके जोड़ें।