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Find ∑r=1n(5r2+4r−3). - Mathematics and Statistics

Sum

Find \[\displaystyle\sum_{r=1}^{n}(5r^2 + 4r - 3)\].

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Solution

\[\displaystyle\sum_{r=1}^{n}(5r^2 + 4r - 3)\]

= 5\[\displaystyle\sum_{r=1}^{n}r^2 + 4\displaystyle\sum_{r=1}^{n} r - 3\displaystyle\sum_{r=1}^{n} 1\]

= `5.("n"("n" + 1)(2"n" + 1))/6 + 4.("n"("n" + 1))/2 - 3"n"`

= `"n"/6[5(2"n"^2 + 3"n" + 1) + 12("n" + 1) - 18]`

= `"n"/6(10"n"^2 + 15"n" + 5 + 12"n" + 12 - 18)`

= `"n"/6(10"n"^2 + 27"n" - 1)`.

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APPEARS IN

Balbharati Mathematics and Statistics 1 (Commerce) 11th Standard Maharashtra State Board
Chapter 4 Sequences and Series
Miscellaneous Exercise 4 | Q 10 | Page 64
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