James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

A36
appendix E Sigma Notation
SOLUTION 1 Let S be the desired sum. We start with the telescoping sum (or collapsing
sum):
Most terms cancel in pairs.
o n
fs1 1 id 3 2 i 3 g − s2 3 2 1 3 d 1 s3 3 2 2 3 d 1 s4 3 2 3 3 d 1 ∙ ∙ ∙ 1 fsn 1 1d 3 2 n 3 g
i−1
− sn 1 1d 3 2 1 3 − n 3 1 3n 2 1 3n
On the other hand, using Theorem 2 and Examples 3 and 4, we have
Thus we have
o n
fs1 1 i d 3 2 i 3 g − o n
f3i 2 1 3i 1 1g − 3 o n
i 2 1 3 o n
i 1 o n
1
i−1
i−1
i−1
i−1 i−1
− 3S 1 3
Solving this equation for S, we obtain
nsn 1 1d
2
n 3 1 3n 2 1 3n − 3S 1 3 2 n 2 1 5 2 n
3S − n 3 1 3 2 n 2 1 1 2 n
1 n − 3S 1 3 2 n 2 1 5 2 n
or
S − 2n 3 1 3n 2 1 n
6
−
nsn 1 1ds2n 1 1d
6
Principle of Mathematical Induction
Let S n be a statement involving the
positive integer n. Suppose that
1. S 1 is true.
2. If S k is true, then S k11 is true.
Then S n is true for all positive integers
n.
See pages 72 and 74 for a more
thorough discussion of mathematical
induction.
SOLUTION 2 Let S n be the given formula.
1. S 1 is true because
1 2 −
2. Assume that S k is true; that is,
Then
1s1 1 1ds2 ? 1 1 1d
6
1 2 1 2 2 1 3 2 1 ∙ ∙ ∙ 1 k 2 −
ksk 1 1ds2k 1 1d
6
1 2 1 2 2 1 3 2 1 ∙ ∙ ∙ 1 sk 1 1d 2 − s1 2 1 2 2 1 3 2 1 ∙ ∙ ∙ 1 k 2 d 1 sk 1 1d 2
So S k11 is true.
−
ksk 1 1ds2k 1 1d
6
− sk 1 1d
1 sk 1 1d 2
ks2k 1 1d 1 6sk 1 1d
6
− sk 1 1d 2k 2 1 7k 1 6
6
−
−
sk 1 1dsk 1 2ds2k 1 3d
6
sk 1 1dfsk 1 1d 1 1gf2sk 1 1d 1 1g
6
By the Principle of Mathematical Induction, S n is true for all n.
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