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दशाइए कि nnnnnn1×22+2×32+....+ n×(n+1)212×2+22×3+....+ n2×(n+1)=3n+53n+1 - Mathematics (गणित)

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Sum

दर्शाइए कि: `(1 xx 2^2 + 2 xx 3^2 + .... +  "n" xx ("n" + 1)^2)/(1^2 xx 2 + 2^2 xx 3 + .... +  "n"^2 xx ("n" + 1)) = (3"n" + 5)/(3"n" + 1)`

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Solution

अंश का nवाँ पद = n(n + 1)2 = n(n2 + 2n + 1)

= n3 + 2n2 + n

अंश के n पदों का योग S1 = `sum"n"^3 + 2sum"n"^2 + sum"n"`

= `("n"^2 ("n" + 1)^2)/4 + (2"n" ("n" + 1) (2"n" + 1))/6 + ("n"("n" + 1))/2`

= `("n"("n" + 1))/12 [3"n" ("n" + 1) + 4(2"n" + 1) + 6]`

= `("n"("n" + 1))/12 [3"n"^2 + 3"n" + 8"n" + 4 + 6]`

= `("n"("n" + 1))/12 [3"n"^2 + 11"n" + 10]`

= `("n"("n" + 1) ("n" + 2) (3"n" + 5))/12`

हर का nवाँ पद = n2 (n + 1) = n3 + n2

हर के n पदों का योग S2 = `sum"n"^3 + sum"n"^2`

= `("n"^2 ("n" + 1)^2)/4 + ("n" ("n" + 1) (2"n" + 1))/6` 

= `("n"("n" + 1))/12 [3"n" ("n" + 1) + 2(2"n" + 1)]`

= `("n"("n" + 1))/12 [3"n"^2 + 3"n" + 4"n" + 2]`

= `("n"("n" + 1))/12 [3"n"^2 + 7"n" + 2]`

= `("n"("n" + 1) ("n" + 2) (3"n" + 1))/12`

`"S"_1/"S"_2 = (("n"("n" + 1) ("n" + 2) (3"n" + 5))/12)/(("n"("n" + 1) ("n" + 2) (3"n" + 1))/12)`

= `(3"n" + 5)/(3"n" + 1)`

Concept: विशेष अनुक्रमों के n पदों का योगफल
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APPEARS IN

NCERT Mathematics Class 11 [गणित कक्षा ११ वीं]
Chapter 9 अनुक्रम तथा श्रेणी
अध्याय 9 पर विविध प्रश्नावली | Q 26. | Page 214
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