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 $\sum_{i=1}^{n} i^{3}=\left[\frac{n(n+1)}{2}\right]^{2}$ to find the sum.
-Find the sum of the first 70 cubed positive integers.
A) $24,700,900$
B) 116,795
C) $12,350,450$
D) $6,175,225$

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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 $\sum_{i=1}^{n} i^{3}=\left[\frac{n(n+1)}{2}\right]^{2}$ to find the sum.
-Find the sum of the first 70 cubed positive integers.
A) $24,700,900$
B) 116,795
C) $12,350,450$
D) $6,175,225$

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 ∑i=1ni3=[n(n+1)2]2\sum_{i=1}^{n} i^{3}=\left[\frac{n(n+1)}{2}\right]^{2}∑i=1n​i3=[2n(n+1)​]2

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 $\sum_{i=1}^{n} i^{3}=\left[\frac{n(n+1)}{2}\right]^{2}$