( \frac{(n+3)!}{n!}+\frac{(n+1)!}{(n-2)!} ) का सरल रूप क्या है?

What is the simplified form of ( \frac{(n+3)!}{n!}+\frac{(n+1)!}{(n-2)!} )?

Explanation opens after your attempt
Correct Answer

A. \(2n^3+6n^2+4n+6\)

Step 1

Concept

The first term is ((n+3)(n+2)(n+1)) and the second is ((n+1)n(n-1)). Adding gives \(2n^3+6n^2+4n+6\).

Step 2

Why this answer is correct

The correct answer is A. \(2n^3+6n^2+4n+6\). The first term is ((n+3)(n+2)(n+1)) and the second is ((n+1)n(n-1)). Adding gives \(2n^3+6n^2+4n+6\).

Step 3

Exam Tip

पहला पद ((n+3)(n+2)(n+1)) और दूसरा ((n+1)n(n-1)) है। जोड़ने पर \(2n^3+6n^2+4n+6\) मिलता है।

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FAQs

Mathematics Answer, Explanation and Revision Hints

( \frac{(n+3)!}{n!}+\frac{(n+1)!}{(n-2)!} ) का सरल रूप क्या है? / What is the simplified form of ( \frac{(n+3)!}{n!}+\frac{(n+1)!}{(n-2)!} )?

Correct Answer: A. \(2n^3+6n^2+4n+6\). Explanation: पहला पद ((n+3)(n+2)(n+1)) और दूसरा ((n+1)n(n-1)) है। जोड़ने पर \(2n^3+6n^2+4n+6\) मिलता है। / The first term is ((n+3)(n+2)(n+1)) and the second is ((n+1)n(n-1)). Adding gives \(2n^3+6n^2+4n+6\).

Which concept should I revise for this Mathematics MCQ?

The first term is ((n+3)(n+2)(n+1)) and the second is ((n+1)n(n-1)). Adding gives \(2n^3+6n^2+4n+6\).

What exam hint can help solve this Mathematics question?

पहला पद ((n+3)(n+2)(n+1)) और दूसरा ((n+1)n(n-1)) है। जोड़ने पर \(2n^3+6n^2+4n+6\) मिलता है।