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If \(\frac{(n+1)!}{(n-1)!}-\frac{n!}{(n-2)!}=24\), what is the value of \(n\)?यदि \(\frac{(n+1)!}{(n-1)!}-\frac{n!}{(n-2)!}=24\), तो \(n\) का मान क्या है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: 1212
Here, \(n\geq 2\). We have \(\frac{(n+1)!}{(n-1)!}=n(n+1)\) and \(\frac{n!}{(n-2)!}=n(n-1)\). Thus, \(n(n+1)-n(n-1)=2n\). Therefore, \(2n=24\), giving \(n=12\). Exam tip: Expand factorial quotients into products of consecutive integers before simplifying.यहाँ \(n\geq 2\) होना चाहिए। \(\frac{(n+1)!}{(n-1)!}=n(n+1)\) और \(\frac{n!}{(n-2)!}=n(n-1)\)। इसलिए \(n(n+1)-n(n-1)=2n\)। अतः \(2n=24\), जिससे \(n=12\)। परीक्षा-युक्ति: फैक्टोरियल के भागों को पहले क्रमागत गुणनफल में बदलें; इससे गणना सरल हो जाती है।
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