( \frac{(n+1)!+n!}{(n-1)!} ) का सरल रूप क्या है?

What is the simplified form of ( \frac{(n+1)!+n!}{(n-1)!} )?

Explanation opens after your attempt
Correct Answer

B. (n(n+2))

Step 1

Concept

((n+1)!+n!=n!((n+1)+1)=n!(n+2)), so the answer is (n(n+2)). First take (n!) as the common factor.

Step 2

Why this answer is correct

The correct answer is B. (n(n+2)). ((n+1)!+n!=n!((n+1)+1)=n!(n+2)), so the answer is (n(n+2)). First take (n!) as the common factor.

Step 3

Exam Tip

((n+1)!+n!=n!((n+1)+1)=n!(n+2)), इसलिए उत्तर (n(n+2)) है। पहले (n!) को सामान्य फैक्टर बनाएं।

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

( \frac{(n+1)!+n!}{(n-1)!} ) का सरल रूप क्या है? / What is the simplified form of ( \frac{(n+1)!+n!}{(n-1)!} )?

Correct Answer: B. (n(n+2)). Explanation: ((n+1)!+n!=n!((n+1)+1)=n!(n+2)), इसलिए उत्तर (n(n+2)) है। पहले (n!) को सामान्य फैक्टर बनाएं। / ((n+1)!+n!=n!((n+1)+1)=n!(n+2)), so the answer is (n(n+2)). First take (n!) as the common factor.

Which concept should I revise for this Mathematics MCQ?

((n+1)!+n!=n!((n+1)+1)=n!(n+2)), so the answer is (n(n+2)). First take (n!) as the common factor.

What exam hint can help solve this Mathematics question?

((n+1)!+n!=n!((n+1)+1)=n!(n+2)), इसलिए उत्तर (n(n+2)) है। पहले (n!) को सामान्य फैक्टर बनाएं।