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1.32 (i) Show, for every n ≥ 1, that the “alternating sum” of the binomial
coefﬁcients is zero:
n
n
n
n
= 0.
−
+
− · · · + (−1)n
n
0
1
2
Solution. If f (x ) = (1 + x )n , then f (−1) = (1 − 1)n = 0; but
the expansion is the alternating sum of the binomial coefﬁcients. (ii) Use part (i) to prove, for a given n , that the sum of all the binomial
n
n
coefﬁcients r with r even is equal to the sum of all those r
with r odd.
Solution. By part (i),
n
n
n
n
−
+
− ··· ±
= 0.
0
1
2
n
Since the sign of n is (−1)r , the terms n with r even are posir
r
tive while those with r odd are negative. Just put those coefﬁcients
with negative coefﬁcient on the other side of the equation.
1.33 Prove that if n ≥ 2, then
n (−1)r −1r r =1 n
= 0.
r Solution. Again, consider f (x ) = (1+ x )n . There are two ways to describe
its derivative f (x ). On the one hand, f (x ) = n (1 + x )n −1 . On the other
hand, we can do termbyterm differentiation:
n f (x ) = r
r =1 n r −1
x.
r Evaluating at x = −1 gives f (−1) = n (1 − 1)n −1 = 0, since n − 1 ≥ 1.
On the other hand, we can use the expansion to see
n f (−1) = (−1)r −1r r =1 1.34 If 1 ≤ r ≤ n , prove that
n n−1
n
=
.
r
r r −1
Solution. Absent. n
.
r ...
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This note was uploaded on 12/21/2011 for the course MAS 4301 taught by Professor Keithcornell during the Fall '11 term at UNF.
 Fall '11
 KeithCornell

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