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Calculus 1 Problems &
Solutions – Chapter 6 – Section 6.1.1
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6.1.1
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Review |
1. Notation |
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Example 1.1
Write out these sums:
EOS
The lower limit of the sum is often 1. It may also be any other non-negative integer, like 0 or 3.
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2. Properties |
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Proof
ii and iii. Similar to i.
EOP
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3. Formulas |
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Proof
i. We use the identity (k + 1)2 – k2 = 2k + 1 (derived from (k + 1)2 = k2
+ 2k
+ 1). Writing it out for each integer k from
1 to n
and adding them up we get:
ii. We use the identity (k + 1)3 – k3
= 3k2 + 3k + 1 (derived from (k + 1)3 = k3
+ 3k2 + 3k + 1). ). Writing it out for each
integer k
from 1 to n
and adding them up we get:
iii and iv. Left as Problems &
Solutions 4 and 5
respectively.
EOP
For formulas i, ii, and iii, the base is increasing from 1
to n
and the exponent is fixed, for example 12
+ 22 + ... + n2, while
for formula iv the base is fixed and the exponent is increasing from 0 to n, for
example 1 + (1/2) + (1/2)2
+ ... + (1/2)n.
Find the following sums.
b. 12 + 22 + ... + 1002.
Solution
EOS
Problems & Solutions |
1. Write out this sum:
Solution
2. Write out this sum:
Solution
3. Write out this sum:
Solution
a. Prove this formula:
Solution
a. Writing the identity (k + 1)4 – k4 = 4k3 +6k2 + 4k + 1 for each integer k from 1 to n and adding them up we get:
Solution
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