Exercises
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Exercises
Write the following sums without the sigma notation then evaluate them.
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Suppose $\sum_{i=1}^n a_n = 1$ and $\sum_{i=1}^n b_n = -3$. Use those values to evaluate the following sums.
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Evaluate the following sums using appropriate formulas
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For each of the following functions complete the following. (a) Graph the function. (b) Divide the given interval into four subintervals and add the corresponding right hand end points. (c) Use the areas of those rectangles to approximate the area under the curve.
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For each of the following functions complete the following. (a) Graph the function. (b) Divide the given interval into four subintervals and add the corresponding left hand end points. (c) Use the areas of those rectangles to approximate the area under the curve.
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For each of the following functions complete the following. (a) Graph the function. (b) Divide the given interval into five subintervals and add the corresponding mid-points. (c) Use the areas of those rectangles to approximate the area under the curve.
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For each of the following functions, find a formula for the Reimann sum formed by dividing the interval into $n$ subintervals using the left-hand end points. Then take the limit as $n\to\infty$ to get a value for the area under the curve.
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Explorations
For each of the following functions, find a formula for the Reimann sum formed by dividing the interval into $n$ subintervals using the left-hand end points. Then take the limit as $n\to\infty$ to get a value for the area under the curve. Explain the results. (They're unusual if you're expecting to get an area. Suggest a solution or modification that will give the physical area under each curve.)
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Write $\sum_{i=1}^3 \frac{1}{n}$ without the sigma notation then evaluate it.
$$\sum_{i=1}^{3} \frac{1}{n} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} = \frac{11}{6}$$
Write $\sum_{i=0}^2 n \sin(n\pi)$ without the sigma notation then evaluate it.
$$\sum_{i=0}^2 n \sin(n\pi) = 0 \sin (0 \pi) + 1 \sin (1 \pi) + 2 \cdot \sin (2 \pi) = 0$$
Write $\sum_{i=1}^4 (-1)^n (n + 1)$ without the sigma notation then evaluate it.
$$\sum_{i=1}^4 (-1)^n (n + 1) = (-1)^0 (0 + 1) + (-1)^1 (1 + 1) + (-1)^2 (2 + 1) + (-1)^3 (3 + 1) = 1 - 2 + 3 + -4 = -2$$
Suppose $\sum_{i=1}^n a_n = 1$ and $\sum_{i=1}^n b_n = -3$. Use those values to evaluate $\sum_{i=1}^n(\frac{a_n}{3})$.
$$\sum_{i=1}^n(\frac{a_n}{3}) = \frac{1}{3} \sum_{i=1}^n a_n = \frac{1}{3} \cdot 1 = \frac{1}{3}$$
Suppose $\sum_{i=1}^n a_n = 1$ and $\sum_{i=1}^n b_n = -3$. Use those values to evaluate $\sum_{i=1}^n(a_n - 3b_n)$.
$$\sum_{i=1}^n(a_n - 3b_n) = \sum_{i=1}^n a_n - 3 \sum_{i=1}^n b_n = 1 - 3 \cdot (-3) = 10 $$
Evaluate $\sum_{n=1}^{10} n^2$.
$$\sum_{n=1}^{10} n^2 = \frac{10(10+1)(2\cdot 10+1)}{6} = 385$$
Evaluate $\sum_{n=1}^4 (2n + n^2)$.
$$\begin{aligned} \sum_{n=1}^4 (2n + n^2) &= 2\sum_{n=1}^4 n + \sum_{n=1}^4 n^2 \\ &= 2\frac{4(4+1)}{2} + \frac{4(4+1)(2\cdot 4+1)}{6} \\ &= 2 \cdot 10 + 30 \\ &= 50 \end{aligned}$$
Evaluate $\sum_{n=1}^{20} n(n - 1)$.
$$\begin{aligned} \sum_{n=1}^{20} n(n - 1) &= \sum_{n=1}^{20} (n^2 - n) \\ &= \sum_{n=1}^{20} n^2 - \sum_{n=1}^{20} n \\ &= \frac{20(20+1)(2\cdot 20+1)}{6} - \frac{20(20 + 1)}{2} \\ &= 2870 - 210 \\ &= 2660 \end{aligned}$$
Evaluate $\sum_{n=6}^{12} 5$.
There's no variable in the expression so this is just saying to add up 12 - 6 + 1 = 7 5's so
$$\sum_{n=6}^{12} 5 = 5 \cdot 7 = 35$$Evaluate $\sum_{n=1}^{k} -3$.
There's no variable in the expression itself so this is just saying to add up k - 1 + 1 = k -3's so
$$\sum_{n=1}^{k} -3 = -3k$$Evaluate $\sum_{n=1}^{k} \frac{n^2}{k}$.
$$\begin{aligned} \sum_{n=1}^{k} \frac{n^2}{k} &= \frac{1}{k}\sum_{n=1}^{k} n^2 \\ &= \frac{1}{k} \frac{n(n+1)(2n + 1)}{6} \\ &= \frac{n(n+1)(2n + 1)}{6k} \end{aligned}$$
