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.