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1. Calculate the sum Pn j=1 j and compare with n(n + 1)/2, for n = 400, 600, 800, 1000. Use the formula to compute Pn j=1 j for all values of n between 1 and 100. 2. Using rep(), : and c()as needed, create the vectors below. To make it harder: in each expression you can only use each function at most twice, and with at most 2 arguments. So c(1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5) is not allowed!) (a) 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 (b) 0 1 1 2 2 2 3 3 3 3 4 4 4 4 4 5 5 5 5 5 5 (c) 1 2 2 3 3 3 1 2 2 3 3 3 1 2 2 3 3 3 (d) "red" "red" "red" "green" "green" "blue" 3. Use R to find the remainder r when dividing 664055 by 243219. Use R to find the whole number part w of the ratio 664055/243219. Use R to confirm that 664055 = 243219w + r. 4. Use R to find the binary expansion of 1/7 to 8 binary places. Let x be the number just computed. By hand using binary arithmetic, compute 2x as x + x. By hand using binary arithmetic, compute 4x as 2x+ 2x. By hand using binary arithmetic, compute 4x+ 2x+x. Using 1 R, compute the decimal representation of your final answer. Explain why you did not get 1. 5. Use the options() function in R to set the number of digits to display to 20. Compute and print 10i + 1 for i = 0, . . . , 20 using a single R statement. Identify and explain cases where R prints the wrong answer. 6. Plot the values 1015+i for i = 1, . . . , 100. (To do this, put the values in a variable x, and run plot(x).) Plot the values 1016+i for i = 1, . . . , 100. Plot the values 1017+i for i = 1, . . . , 100. Explain why these three plots look so different, and why each one looks the way it does.

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### 1.
```{r}
sum(1:400)
sum(1:600)
sum(1:800)
sum(1:1000)
n <- 1:100
n*(n+1)/2
```

### 2.
#### (a)
```{r}
v <- rep(1:5,4)
```

#### (b)
```{r}
v <- rep(1:6,1:6)-1
```

### (c)
```{r}
v <- rep( rep(1:3,1:3), 3)

```
### (d)
```{r}
v <- rep(c("red","green","blue"),3:1)...

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