किसी अनुक्रम में \(P_1=0\) और \(P_n=P_{n-1}+3n^2-1\) है। \(P_4\) का मान क्या होगा?

In a sequence, \(P_1=0\) and \(P_n=P_{n-1}+3n^2-1\). What is the value of \(P_4\)?

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

C. चौरासी(84)

Step 1

Concept

\(P_4=0+11+26+47=84\). In exams, calculate \(3n^2-1\) first.

Step 2

Why this answer is correct

The correct answer is C. चौरासी / (84). \(P_4=0+11+26+47=84\). In exams, calculate \(3n^2-1\) first.

Step 3

Exam Tip

\(P_4=0+11+26+47=84\) है। परीक्षा में \(3n^2-1\) को पहले निकालें।

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

किसी अनुक्रम में \(P_1=0\) और \(P_n=P_{n-1}+3n^2-1\) है। \(P_4\) का मान क्या होगा? / In a sequence, \(P_1=0\) and \(P_n=P_{n-1}+3n^2-1\). What is the value of \(P_4\)?

Correct Answer: C. चौरासी / (84). Explanation: \(P_4=0+11+26+47=84\) है। परीक्षा में \(3n^2-1\) को पहले निकालें। / \(P_4=0+11+26+47=84\). In exams, calculate \(3n^2-1\) first.

Which concept should I revise for this Mathematics MCQ?

\(P_4=0+11+26+47=84\). In exams, calculate \(3n^2-1\) first.

What exam hint can help solve this Mathematics question?

\(P_4=0+11+26+47=84\) है। परीक्षा में \(3n^2-1\) को पहले निकालें।