Question
Transcribed Text
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 k1 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 rewrite j back to k. Find that it is equal to
n
(1  x)S = k=1 Xk 
(e)
xn1
The sigma notation in the right hand side is X
as we remember in class. Solve this for S to
x1
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 1a.
1a. 1
1b. \binom(2n + 1)(n + 1)
1c. n \binom(2n  1)(n  1)
1d. 1
2a. x^(k + 1)
2b. 2
2c. n + 1
2d. (j  1)x^j
2e. nx^(n + 1)  x^(n + 1)
3a. \ln k
3b. 0
3c. \ln (n  1)
3d. 1
3e. (n  1)^(n  1)...
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