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If 1×2+2×3+3×4+4×5+... upto n terms1+2+3+4+... upto n terms=1003, find n - Mathematics and Statistics

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प्रश्न

If `(1 xx 2 + 2 xx 3 + 3 xx 4 + 4 xx 5 + ...  "upto n terms")/(1 + 2 + 3 + 4 + ...  "upto n terms") = 100/3,` find n

योग

उत्तर

`(1 xx 2 + 2 xx 3 + 3 xx 4 + 4 xx 5 + ...  "upto n terms")/(1 + 2 + 3 + 4 + ...  "upto n terms") = 100/3,`

∴ `(sum_("r" = 1)^"n""r"("r" + 1))/(sum_("r" = 1)^"n" "r") = 100/3`

∴ `(sum_("r" = 1)^"n" "r"^2 + sum_("r" = 1)^"n" "r")/(sum_("r" = 1)^"n" "r") = 100/3`

∴ `(("n"("n" + 1)(2"n" + 1))/6 + ("n"("n" + 1))/2)/(("n"("n" + 1))/2) = 100/3`

∴ `(("n"("n" + 1))/6[(2"n" + 1) + 3])/(("n"("n" + 1))/2) = 100/3`

∴ `(2("n" + 2))/3= 100/3`

∴ n + 2 = 50

∴  n = 48

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Arithmetico Geometric Series
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 2: Sequences and Series - Exercise 2.6 [पृष्ठ ४०]

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बालभारती Mathematics and Statistics 2 (Arts and Science) [English] 11 Standard Maharashtra State Board
अध्याय 2 Sequences and Series
Exercise 2.6 | Q 9 | पृष्ठ ४०

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