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MISCELLANEOUS EXERCISE - 2 (II) · Q113

Q.Find ∑r=1n(5r2+4r−3)\displaystyle\sum_{r=1}^{n}(5r^2+4r-3).

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∑(5r2+4r−3)=5⋅n(n+1)(2n+1)6+4⋅n(n+1)2−3n=5n(n+1)(2n+1)6+2n(n+1)−3n\sum(5r^2+4r-3)=5\cdot\dfrac{n(n+1)(2n+1)}6+4\cdot\dfrac{n(n+1)}2-3n=\dfrac{5n(n+1)(2n+1)}6+2n(n+1)-3n. …

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