Use the Formulas$\sum_{i=1}^{n} i=\frac{n(n+1)}{2}, \sum_{i=1}^{n} i^{2}=\frac{n(n+1)(2 n+1)}{6}$ , And$\sum_{i=1}^{n}
-\(\sum_{i=1}^{10}\left(2 i^{3}-5 i^{2}\right)$

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Use the Formulas$\sum_{i=1}^{n} i=\frac{n(n+1)}{2}, \sum_{i=1}^{n} i^{2}=\frac{n(n+1)(2 n+1)}{6}$ , And$\sum_{i=1}^{n} 
-\(\sum_{i=1}^{10}\left(2 i^{3}-5 i^{2}\right)$

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The Answer of Use the Formulas$\sum_{i=1}^{n} i=\frac{n(n+1)}{2}, \sum_{i=1}^{n} i^{2}=\frac{n(n+1)(2 n+1)}{6}$ , And$\sum_{i=1}^{n} 
-\(\sum_{i=1}^{10}\left(2 i^{3}-5 i^{2}\right)$

Use the Formulas $\sum_{i=1}^{n} i=\frac{n(n+1)}{2}, \sum_{i=1}^{n} i^{2}=\frac{n(n+1)(2 n+1)}{6}$ , And $\sum_{i=1}^{n} -\(\sum_{i=1}^{10}\left(2 i^{3}-5 i^{2}\right)$

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Use the Formulas ∑i=1ni=n(n+1)2,∑i=1ni2=n(n+1)(2n+1)6\sum_{i=1}^{n} i=\frac{n(n+1)}{2}, \sum_{i=1}^{n} i^{2}=\frac{n(n+1)(2 n+1)}{6}∑i=1n​i=2n(n+1)​,∑i=1n​i2=6n(n+1)(2n+1)​ , And \sum_{i=1}^{n} -\(\sum_{i=1}^{10}\left(2 i^{3}-5 i^{2}\right)

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Use the Formulas $\sum_{i=1}^{n} i=\frac{n(n+1)}{2}, \sum_{i=1}^{n} i^{2}=\frac{n(n+1)(2 n+1)}{6}$ , And $\sum_{i=1}^{n} -\(\sum_{i=1}^{10}\left(2 i^{3}-5 i^{2}\right)$

Correct Answer:
Verified
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Get Access to more Verified Answers free of charge