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Simplify n2-9(n+3)!+6(n+2)!-1(n+1)!

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प्रश्न

Simplify `("n"^2 - 9)/(("n" + 3)!) + 6/(("n" + 2)!) - 1/(("n" + 1)!)`

योग
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उत्तर

`("n"^2 - 9)/(("n" + 3)!) + 6/(("n" + 2)!) - 1/(("n" + 1)!)`

= `(("n" - 3)("n" + 3))/(("n" + 3)("n" + 2)("n" + 1)!) + 6/(("n" + 2)("n" + 1)!) - 1/(("n" + 1)!)`

= `1/(("n" + 1)!)[("n" - 3)/("n" + 2) + 6/("n" + 2) - 1]`

= `1/(("n" + 1)!)[("n" - 3 + 6 - "n" - 2)/("n" + 2)]`

= `1/(("n" + 2)("n" + 1)!)`

= `1/(("n" + 2)!)`

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अध्याय 3: Permutations and Combination - Exercise 3.2 [पृष्ठ ५०]

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बालभारती Mathematics and Statistics (Arts and Science) Part 2 [English] Standard 11 Maharashtra State Board
अध्याय 3 Permutations and Combination
Exercise 3.2 | Q 10. (viii) | पृष्ठ ५०

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