Finding the Value of the Alternating Series 1 - 2 + 3 - 4 + ... + 101

The problem asks us to find the sum of the series \(1 - 2 + 3 - 4 + 5 - \dots\) up to + 101. This is an alternating series where the signs of the terms switch between positive and negative.

The series can be written as: \(1 + (-2) + 3 + (-4) + 5 + (-6) + \dots + 99 + (-100) + 101\).

We can group the terms in pairs to simplify the calculation. Let's group consecutive terms:

\((1 - 2) + (3 - 4) + (5 - 6) + \dots\)

Notice that each pair \((n - (n+1))\) where \(n\) is an odd number gives a sum of -1. For example:

• 1 - 2 = -1
• 3 - 4 = -1
• 5 - 6 = -1

The series goes up to +101. The pairs involving even numbers are \((1, 2), (3, 4), \dots, (99, 100)\). The term +101 is left unpaired at the end.

Let's identify the pairs:

\((1 - 2), (3 - 4), (5 - 6), \dots, (99 - 100)\).

The first number in each pair is 1, 3, 5, ..., 99. This is an arithmetic sequence with the first term \(a_1 = 1\), common difference \(d = 2\), and the last term \(a_n = 99\).

To find the number of terms (\(n\)) in this sequence (which is also the number of pairs), we use the formula: \(a_n = a_1 + (n-1)d\).

Substituting the values:

\(99 = 1 + (n-1)2\)

\(99 - 1 = (n-1)2\)

\(98 = (n-1)2\)

\(\frac{98}{2} = n-1\)

\(49 = n-1\)

\(n = 49 + 1\)

\(n = 50\)

So, there are 50 such pairs, and each pair sums to -1.

The sum of these 50 pairs is \(50 \times (-1) = -50\).

The original series is the sum of these pairs plus the last term, 101:

Sum \(= (1 - 2) + (3 - 4) + \dots + (99 - 100) + 101\)

Sum \(= (-1) + (-1) + \dots + (-1) \text{ (50 times)} + 101\)

Sum \(= 50 \times (-1) + 101\)

Sum = -50 + 101

Sum = 51

Alternatively, we can group the terms differently:

\(1 + (-2 + 3) + (-4 + 5) + (-6 + 7) + \dots + (-100 + 101)\).

Notice that each pair \((-n + (n+1))\) where \(n\) is an even number gives a sum of +1. For example:

• -2 + 3 = 1
• -4 + 5 = 1
• -6 + 7 = 1

The initial term is 1. The pairs are formed by \((-2+3), (-4+5), \dots, (-100+101)\).

The first number in each bracketed pair is -2, -4, -6, ..., -100. The positive counterparts are 2, 4, 6, ..., 100. This is an arithmetic sequence with \(a_1 = 2\), \(d = 2\), and \(a_n = 100\).

To find the number of terms (\(n\)) in this sequence (which is the number of pairs), we use the formula: \(a_n = a_1 + (n-1)d\).

Substituting the values:

\(100 = 2 + (n-1)2\)

\(100 - 2 = (n-1)2\)

\(98 = (n-1)2\)

\(\frac{98}{2} = n-1\)

\(49 = n-1\)

\(n = 49 + 1\)

\(n = 50\)

So, there are 50 such pairs, and each pair sums to +1.

The original series is the sum of the first term plus the sum of these pairs:

Sum \(= 1 + (-2 + 3) + (-4 + 5) + \dots + (-100 + 101)\)

Sum \(= 1 + (+1) + (+1) + \dots + (+1) \text{ (50 times)}\)

Sum \(= 1 + 50 \times (+1)\)

Sum = 1 + 50

Sum = 51

Both grouping methods give the same result, 51.

Thus, the value of the series \(1 - 2 + 3 - 4 + 5 - \dots + 101\) is 51.

Revision Table: Alternating Series Sum

Additional Information: Summation Notations

An alternating series can often be represented using summation notation (\(\Sigma\)). The given series is \(1 - 2 + 3 - 4 + \dots + 101\).

The general term \(a_k\) depends on whether \(k\) is odd or even.

• If \(k\) is odd, \(a_k = +k\).
• If \(k\) is even, \(a_k = -k\).

This can be written using \((-1)^{k+1}k\) or similar expressions.

• For \(k=1\): \((-1)^{1+1} \times 1 = (-1)^2 \times 1 = 1 \times 1 = 1\)
• For \(k=2\): \((-1)^{2+1} \times 2 = (-1)^3 \times 2 = -1 \times 2 = -2\)
• For \(k=3\): \((-1)^{3+1} \times 3 = (-1)^4 \times 3 = 1 \times 3 = 3\)
• ...
• For \(k=101\): \((-1)^{101+1} \times 101 = (-1)^{102} \times 101 = 1 \times 101 = 101\)

So the series can be written as \(\sum_{k=1}^{101} (-1)^{k+1}k\).

Calculating the sum directly using this formula can be complex without recognizing the pairing pattern. The grouping method demonstrated in the solution is a common technique for solving such alternating series problems.