किसी अनुक्रम में \(C_1=4\) और (C_n=C_{n-1}+n(n+2)) है। \(C_4\) का मान क्या होगा?

In a sequence, \(C_1=4\) and (C_n=C_{n-1}+n(n+2)). What is the value of \(C_4\)?

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

C. इक्यावन(51)

Step 1

Concept

\(C_4=4+8+15+24=51\). In exams, simplify (n(n+2)) first.

Step 2

Why this answer is correct

The correct answer is C. इक्यावन / (51). \(C_4=4+8+15+24=51\). In exams, simplify (n(n+2)) first.

Step 3

Exam Tip

\(C_4=4+8+15+24=51\) है। परीक्षा में (n(n+2)) को पहले सरल करें।

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 \(C_1=4\) और (C_n=C_{n-1}+n(n+2)) है। \(C_4\) का मान क्या होगा? / In a sequence, \(C_1=4\) and (C_n=C_{n-1}+n(n+2)). What is the value of \(C_4\)?

Correct Answer: C. इक्यावन / (51). Explanation: \(C_4=4+8+15+24=51\) है। परीक्षा में (n(n+2)) को पहले सरल करें। / \(C_4=4+8+15+24=51\). In exams, simplify (n(n+2)) first.

Which concept should I revise for this Mathematics MCQ?

\(C_4=4+8+15+24=51\). In exams, simplify (n(n+2)) first.

What exam hint can help solve this Mathematics question?

\(C_4=4+8+15+24=51\) है। परीक्षा में (n(n+2)) को पहले सरल करें।