34. Sum of series problems

34. Sum of series problems - step by step solution procedure and detailed explanation for each problem.

Below are 10 sum of series problems, covering arithmetic series, geometric series, and other types of series (e.g., telescoping, harmonic, and special series). Each problem includes a step-by-step solution procedure and a detailed explanation of the reasoning, formulas, and techniques used. The problems range from basic to moderately advanced to provide a comprehensive understanding of summing series.

Problem 1: Sum of an Arithmetic Series

Problem: Find the sum of the first 10 terms of the arithmetic series: 2, 5, 8, 11, ...

Solution Procedure:

Identify the series type: Arithmetic, as the difference between consecutive terms is constant.
Find the common difference (( d )):
5 - 2 = 3
, so
d = 3
.
Identify the first term (
a_1
):
a_1 = 2
.
Determine the number of terms (( n )):
n = 10
.
Find the 10th term (
a_{10}
): Use
a_n = a_1 + (n-1)d
.
- a_{10} = 2 + (10-1) \cdot 3 = 2 + 9 \cdot 3 = 2 + 27 = 29
  .
Use the arithmetic series sum formula:
S_n = \frac{n}{2}(a_1 + a_n)
.
- S_{10} = \frac{10}{2}(2 + 29) = 5 \cdot 31 = 155
  .
Alternative formula:
S_n = \frac{n}{2}[2a_1 + (n-1)d]
.
- S_{10} = \frac{10}{2}[2 \cdot 2 + (10-1) \cdot 3] = 5 \cdot [4 + 9 \cdot 3] = 5 \cdot [4 + 27] = 5 \cdot 31 = 155
  .
State the answer: The sum is 155.

Detailed Explanation:

Arithmetic series: The sum of terms in an arithmetic sequence, where each term increases by a constant difference.
Sum formula:
S_n = \frac{n}{2}(a_1 + a_n)
averages the first and last terms and multiplies by the number of terms. Alternatively,
S_n = \frac{n}{2}[2a_1 + (n-1)d]
uses the first term and common difference.
Verification: List terms: 2, 5, 8, 11, 14, 17, 20, 23, 26, 29. Sum:
2 + 5 + 8 + 11 + 14 + 17 + 20 + 23 + 26 + 29 = 155
.
Formula:
S_n = \frac{n}{2}(a_1 + a_n)
or
S_n = \frac{n}{2}[2a_1 + (n-1)d]
.

Answer: 155.

Problem 2: Sum of a Geometric Series

Problem: Find the sum of the first 6 terms of the geometric series: 3, 6, 12, 24, ...

Solution Procedure:

Identify the series type: Geometric, as each term is multiplied by a constant ratio.
Find the common ratio (( r )):
\frac{6}{3} = 2
, so
r = 2
.
Identify the first term (( a )):
a = 3
.
Determine the number of terms (( n )):
n = 6
.
Use the geometric series sum formula: For
r \neq 1
,
S_n = a \frac{1 - r^n}{1 - r}
.
- S_6 = 3 \frac{1 - 2^6}{1 - 2}
  .
Calculate:
- 2^6 = 64
  , so
  1 - 2^6 = 1 - 64 = -63
  .
- 1 - 2 = -1
  .
- S_6 = 3 \cdot \frac{-63}{-1} = 3 \cdot 63 = 189
  .
State the answer: The sum is 189.

Detailed Explanation:

Geometric series: The sum of terms in a geometric sequence, where each term is the previous term multiplied by a constant ratio.
Sum formula:
S_n = a \frac{1 - r^n}{1 - r}
accounts for the geometric progression. The numerator
1 - r^n
is the difference of a geometric sequence, and the denominator adjusts for the ratio.
Verification: Terms: 3, 6, 12, 24, 48, 96. Sum:
3 + 6 + 12 + 24 + 48 + 96 = 189
.
Formula:
S_n = a \frac{1 - r^n}{1 - r}
.

Answer: 189.

Problem 3: Sum of an Arithmetic Series with Given Terms

Problem: The 3rd term of an arithmetic sequence is 10, and the 8th term is 25. Find the sum of the first 12 terms.

Solution Procedure:

Use the arithmetic nth term formula:
a_n = a_1 + (n-1)d
.
Set up equations for given terms:
- For
  n = 3
  :
  a_3 = a_1 + 2d = 10
  .
- For
  n = 8
  :
  a_8 = a_1 + 7d = 25
  .
Solve the system:
- Subtract:
  (a_1 + 7d) - (a_1 + 2d) = 25 - 10
  , so
  5d = 15
  ,
  d = 3
  .
- Substitute
  d = 3
  into
  a_1 + 2d = 10
  :
  a_1 + 2 \cdot 3 = 10
  ,
  a_1 + 6 = 10
  ,
  a_1 = 4
  .
Find the 12th term:
a_{12} = a_1 + (12-1)d = 4 + 11 \cdot 3 = 4 + 33 = 37
.
Use the sum formula:
S_n = \frac{n}{2}(a_1 + a_n)
.
- S_{12} = \frac{12}{2}(4 + 37) = 6 \cdot 41 = 246
  .
State the answer: The sum is 246.

Detailed Explanation:

Given terms: Use the nth term formula to find
a_1
and ( d ), then compute the last term needed for the sum.
System solving: Two equations solve for two unknowns, ensuring the sequence is fully defined.
Verification: Sequence starts: 4, 7, 10, 13, ..., with
a_8 = 25
,
a_{12} = 37
. Sum formula is efficient for arithmetic series.
Formula:
S_n = \frac{n}{2}(a_1 + a_n)
.

Answer: 246.

Problem 4: Sum of a Geometric Series with Fractional Ratio

Problem: Find the sum of the first 5 terms of the geometric series: 16, 8, 4, 2, ...

Solution Procedure:

Identify the series type: Geometric, as each term is multiplied by
\frac{1}{2}
.
Find the common ratio:
\frac{8}{16} = \frac{1}{2}
, so
r = \frac{1}{2}
.
Identify the first term:
a = 16
.
Determine the number of terms:
n = 5
.
Use the geometric sum formula:
S_n = a \frac{1 - r^n}{1 - r}
.
- S_5 = 16 \frac{1 - \left(\frac{1}{2}\right)^5}{1 - \frac{1}{2}}
  .
Calculate:
- \left(\frac{1}{2}\right)^5 = \frac{1}{32}
  , so
  1 - \frac{1}{32} = \frac{31}{32}
  .
- 1 - \frac{1}{2} = \frac{1}{2}
  .
- S_5 = 16 \cdot \frac{\frac{31}{32}}{\frac{1}{2}} = 16 \cdot \frac{31}{32} \cdot 2 = 16 \cdot \frac{31}{16} = 31
  .
State the answer: The sum is 31.

Detailed Explanation:

Fractional ratio: A ratio
|r| < 1
means terms decrease, but the sum formula applies the same way.
Simplification: The fraction division simplifies to multiplication by the reciprocal.
Verification: Terms: 16, 8, 4, 2, 1. Sum:
16 + 8 + 4 + 2 + 1 = 31
.
Formula:
S_n = a \frac{1 - r^n}{1 - r}
.

Answer: 31.

Problem 5: Sum of a Telescoping Series

Problem: Find the sum of the series:

\sum_{k=1}^{10} \left( \frac{1}{k} - \frac{1}{k+1} \right)

Solution Procedure:

Identify the series type: Telescoping, as terms may cancel when summed.
Write out the series:
- \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \cdots + \left( \frac{1}{10} - \frac{1}{11} \right)
  .
Sum the terms:
- Most terms cancel:
  \frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \cdots + \frac{1}{10} - \frac{1}{11}
  .
- Remaining:
  \frac{1}{1} - \frac{1}{11} = 1 - \frac{1}{11} = \frac{10}{11}
  .
State the answer: The sum is
\frac{10}{11}
.

Detailed Explanation:

Telescoping series: Terms pair up to cancel, leaving only the first and last parts.
General form: For
\sum_{k=1}^n \left( \frac{1}{k} - \frac{1}{k+1} \right)
, the sum is
1 - \frac{1}{n+1}
.
Verification: Compute for small ( n ): For
n = 2
,
\left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) = 1 - \frac{1}{3} = \frac{2}{3}
.
Formula: Sum collapses to
1 - \frac{1}{n+1}
.

Answer:

\frac{10}{11}

Problem 6: Sum of the First n Natural Numbers

Problem: Find the sum of the series:

1 + 2 + 3 + \cdots + 50

Solution Procedure:

Identify the series type: Arithmetic series of the first ( n ) natural numbers.
Determine parameters:
- First term (
  a_1
  ): 1.
- Last term (
  a_n
  ): 50.
- Number of terms (( n )): 50.
Use the sum formula:
S_n = \frac{n}{2}(a_1 + a_n)
.
- S_{50} = \frac{50}{2}(1 + 50) = 25 \cdot 51 = 1275
  .
Alternative formula: Sum of first ( n ) natural numbers:
S_n = \frac{n(n+1)}{2}
.
- S_{50} = \frac{50 \cdot 51}{2} = \frac{2550}{2} = 1275
  .
State the answer: The sum is 1275.

Detailed Explanation:

Natural numbers: A special arithmetic series with
a_1 = 1
,
d = 1
.
Special formula:
\frac{n(n+1)}{2}
is derived from pairing terms (e.g.,
1 + n
,
2 + (n-1)
) or Gaussian summation.
Verification: For small ( n ):
n = 3
,
1 + 2 + 3 = 6
,
\frac{3 \cdot 4}{2} = 6
.
Formula:
S_n = \frac{n(n+1)}{2}
.

Answer: 1275.

Problem 7: Sum of a Finite Series with Linear Terms

Problem: Find the sum:

\sum_{k=1}^{8} (2k + 3)

Solution Procedure:

Expand the general term:
2k + 3
.
Write the series:
(2 \cdot 1 + 3) + (2 \cdot 2 + 3) + \cdots + (2 \cdot 8 + 3) = 5 + 7 + 9 + \cdots + 19
.
Identify as arithmetic:
- First term: 5.
- Common difference:
  7 - 5 = 2
  .
- Number of terms: 8.
Find the 8th term:
a_8 = 2 \cdot 8 + 3 = 19
.
Use the sum formula:
S_n = \frac{n}{2}(a_1 + a_n)
.
- S_8 = \frac{8}{2}(5 + 19) = 4 \cdot 24 = 96
  .
Alternative method: Split the sum:
- \sum_{k=1}^8 (2k + 3) = 2 \sum_{k=1}^8 k + \sum_{k=1}^8 3
  .
- Sum of ( k ):
  \sum_{k=1}^8 k = \frac{8 \cdot 9}{2} = 36
  .
- Sum of constant:
  3 \cdot 8 = 24
  .
- Total:
  2 \cdot 36 + 24 = 72 + 24 = 96
  .
State the answer: The sum is 96.

Detailed Explanation:

Linear terms: The general term
2k + 3
forms an arithmetic sequence.
Splitting sums: Separating the linear and constant parts leverages known formulas.
Verification: Sum terms: 5, 7, 9, 11, 13, 15, 17, 19. Total:
5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 96
.
Formula: Arithmetic sum or sum splitting.

Answer: 96.

Problem 8: Sum of a Geometric Series with Negative Ratio

Problem: Find the sum of the first 4 terms of the series: 1, -3, 9, -27, ...

Solution Procedure:

Identify the series type: Geometric, with alternating signs.
Find the common ratio:
\frac{-3}{1} = -3
,
\frac{9}{-3} = -3
, so
r = -3
.
Identify the first term:
a = 1
.
Determine the number of terms:
n = 4
.
Use the geometric sum formula:
S_n = a \frac{1 - r^n}{1 - r}
.
- S_4 = 1 \cdot \frac{1 - (-3)^4}{1 - (-3)}
  .
Calculate:
- (-3)^4 = 81
  , so
  1 - 81 = -80
  .
- 1 - (-3) = 1 + 3 = 4
  .
- S_4 = \frac{-80}{4} = -20
  .
State the answer: The sum is -20.

Detailed Explanation:

Negative ratio: The formula holds for any
r \neq 1
, including negative values.
Alternating terms: The negative ratio causes signs to alternate, affecting the sum.
Verification: Terms: 1, -3, 9, -27. Sum:
1 + (-3) + 9 + (-27) = 1 - 3 + 9 - 27 = -20
.
Formula:
S_n = a \frac{1 - r^n}{1 - r}
.

Answer: -20.

Problem 9: Sum of Squares

Problem: Find the sum:

1^2 + 2^2 + 3^2 + \cdots + 10^2

Solution Procedure:

Identify the series: Sum of the squares of the first ( n ) natural numbers.
Determine the number of terms:
n = 10
.
Use the sum of squares formula:
\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}
.
Substitute:
n = 10
.
- \sum_{k=1}^{10} k^2 = \frac{10 \cdot 11 \cdot 21}{6}
  .
Calculate:
- 10 \cdot 11 = 110
  ,
  110 \cdot 21 = 2310
  .
- \frac{2310}{6} = 385
  .
State the answer: The sum is 385.

Detailed Explanation:

Sum of squares: A known formula for summing
k^2
, derived from polynomial expansion or induction.
Verification: Compute:
1^2 + 2^2 + 3^2 + \cdots + 10^2 = 1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100 = 385
.
Why this formula? It accounts for the quadratic growth of terms.
Formula:
\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}
.

Answer: 385.

Problem 10: Sum of a Harmonic Series

Problem: Find the sum:

\sum_{k=1}^5 \frac{1}{k} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}

Solution Procedure:

Identify the series: Harmonic series, summing the reciprocals of the first ( n ) natural numbers.
Determine the number of terms:
n = 5
.
Compute the sum directly (no closed form for finite harmonic series):
- \frac{1}{1} = 1
  .
- \frac{1}{2} = 0.5
  .
- \frac{1}{3} \approx 0.3333
  .
- \frac{1}{4} = 0.25
  .
- \frac{1}{5} = 0.2
  .
Add the terms:
- Find a common denominator (LCM of 1, 2, 3, 4, 5 is 60):
  - \frac{1}{1} = \frac{60}{60}
    .
  - \frac{1}{2} = \frac{30}{60}
    .
  - \frac{1}{3} = \frac{20}{60}
    .
  - \frac{1}{4} = \frac{15}{60}
    .
  - \frac{1}{5} = \frac{12}{60}
    .
- Sum:
  \frac{60 + 30 + 20 + 15 + 12}{60} = \frac{137}{60}
  .
State the answer: The sum is
\frac{137}{60}
.

Detailed Explanation:

Harmonic series: No simple closed form for finite sums, so we compute directly.
Common denominator: Ensures exact addition of fractions.
Verification: Numeric sum:
1 + 0.5 + 0.3333 + 0.25 + 0.2 \approx 2.2833
, and
\frac{137}{60} \approx 2.2833
.
Formula: None for finite harmonic sums; compute term by term.

Answer:

\frac{137}{60}

Summary of Key Concepts

Arithmetic series: Sum using
S_n = \frac{n}{2}(a_1 + a_n)
or
S_n = \frac{n}{2}[2a_1 + (n-1)d]
.
Geometric series: Sum using
S_n = a \frac{1 - r^n}{1 - r}
for
r \neq 1
.
Telescoping series: Identify cancellations to simplify the sum.
Special series: Use formulas like
\sum k = \frac{n(n+1)}{2}
,
\sum k^2 = \frac{n(n+1)(2n+1)}{6}
.
Harmonic series: Sum reciprocals directly for finite terms.
Splitting sums: Break complex terms into simpler series (e.g., linear terms).
Verification: Manual summation or checking small cases ensures accuracy.

These problems cover a variety of series types and summation techniques, with detailed explanations to clarify the process.

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34. Sum of series problems

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