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1. (2 X 4=8 pts) Find the shortest possible closed formula equal to each. X - Allowed key stroke: binomial coefficient : \binom(x)(y) y a. n 1 - n 2 + n 3 - + (-1)n+1 - n n b. 0 n 1 + 2 n ( 2n + ) n Hint: Write it down with the sigma notation. Use two binomial coefficient identities learned in class. Which are they? n 2 n 2 n 2 n 2 c. 1 + 2 + 3 + + n 1 2 3 n Hint: Use three learned binomial coefficient identities. d. (") - 1 ) 1 n - m n - 1 ) m - n 1 n j - ) ( - j - 1 m - 1 ) 2. (5pts) Below it is discussed how to simplify k-1 kxk (x 1) with change of index. Fill in the five blanks with the shortest possible expressions to complete the sentences. Allowed keystroke: power 'N' Put the sum in S. Consider the two sums in n n S - X S = kxk - k (a) k=1 k=1 In the second sigma notation, change the index k into j=k+1. So we have n (c) (1 S = - (d) k=1 Then re-write j back to k. Find that it is equal to n (1 - x)S = k=1 Xk - (e) xn-1 The sigma notation in the right hand side is X as we remember in class. Solve this for S to x-1 find the closed formula. 3. (7pts) Fill in the five blanks to complete the argument below to approximate n! as a double inequality. (a), (b), (c): 1pt, (d), (e): 2pts Allowed key strokes: natural logarithm base'e', In '\ln' Let t = Inn! where In X means the natural logarithm of X. Observe n t = (a) k=1 Noting f (x) = In x is monotonically increasing, find with the telescope method that t > (b) That is, use the fact that the area of the red boxes is greater than the area between f(x) and x=0. f(c) = In x x 1 2 3 4 n Then shift the red boxes to the right by distance 1, removing the last one. Now the area is smaller than that between f(x) and x=0, so t<__(c). As a result, (d) < n! < (e) achieving our goal.

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#   - Write the answer in each line never changing the labels such as 1-a.


1-a. 1
1-b. \binom(2n + 1)(n + 1)
1-c. n \binom(2n - 1)(n - 1)
1-d. 1


2-a. x^(k + 1)
2-b. 2
2-c. n + 1
2-d. (j - 1)x^j
2-e. nx^(n + 1) - x^(n + 1)


3-a. \ln k
3-b. 0
3-c. \ln (n - 1)
3-d. 1
3-e. (n - 1)^(n - 1)...

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