यदि ( \frac{(n+2)!}{n!}+\frac{n!}{(n-2)!}=114 ), तो (n) का मान क्या है?

If ( \frac{(n+2)!}{n!}+\frac{n!}{(n-2)!}=114 ), what is the value of (n)?

Explanation opens after your attempt
Correct Answer

B. (7)

Step 1

Concept

The simplified form is \(2n^2+2n+2\). Putting (n=7) gives (98+14+2=114).

Step 2

Why this answer is correct

The correct answer is B. (7). The simplified form is \(2n^2+2n+2\). Putting (n=7) gives (98+14+2=114).

Step 3

Exam Tip

सरल रूप \(2n^2+2n+2\) है। (n=7) रखने पर (98+14+2=114) मिलता है।

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Mathematics Answer, Explanation and Revision Hints

यदि ( \frac{(n+2)!}{n!}+\frac{n!}{(n-2)!}=114 ), तो (n) का मान क्या है? / If ( \frac{(n+2)!}{n!}+\frac{n!}{(n-2)!}=114 ), what is the value of (n)?

Correct Answer: B. (7). Explanation: सरल रूप \(2n^2+2n+2\) है। (n=7) रखने पर (98+14+2=114) मिलता है। / The simplified form is \(2n^2+2n+2\). Putting (n=7) gives (98+14+2=114).

Which concept should I revise for this Mathematics MCQ?

The simplified form is \(2n^2+2n+2\). Putting (n=7) gives (98+14+2=114).

What exam hint can help solve this Mathematics question?

सरल रूप \(2n^2+2n+2\) है। (n=7) रखने पर (98+14+2=114) मिलता है।