A sequence is a function defined on the natural numbers: to each index $n = 1, 2, 3, \dots$ it assigns a term $a_n$. The formula for $a_n$ is called the general term. Two basic skills are tested on YÖS:
1) Substitution. To find a specific term, substitute the index into the formula. If $(a_n) = (3n+7)$, then $a_{n+1} = 3(n+1) + 7 = 3n + 10$. Note: you replace every occurrence of $n$ by $n+1$.
2) Solving for the index. If the value of a term is known, set the general term equal to it and solve for the index. If $(a_n) = \left(\frac{2n+1}{3n+4}\right)$ and $a_k = \frac{1}{2}$, then $\frac{2k+1}{3k+4} = \frac{1}{2}$, so $2(2k+1) = 3k+4$, i.e. $4k+2 = 3k+4$, hence $k = 2$. Always check that the answer is a natural number.
The symbol $\sum_{k=m}^{n} a_k$ means the sum $a_m + a_{m+1} + \dots + a_n$: the index $k$ runs through all integers from the lower bound $m$ to the upper bound $n$, and the terms are added.
Number of terms: $n - m + 1$. For example, $\sum_{k=7}^{20} a_k$ has $20 - 7 + 1 = 14$ terms.
The letter used for the index does not matter: $\sum_{k=1}^{n} a_k = \sum_{i=1}^{n} a_i$. Watch the lower bound carefully — sums often start at $k = 0$, not $k = 1$. For instance $\sum_{k=0}^{8} 2^k$ has $9$ terms: $2^0 + 2^1 + \dots + 2^8$.
Constant: a constant $c$ added $n - m + 1$ times gives
$$\sum_{k=m}^{n} c = c \cdot (n - m + 1)$$
Example: $\sum_{k=1}^{5} 2 = 2 \cdot 5 = 10$.
Linearity:
$$\sum_{k=m}^{n} (a_k + b_k) = \sum_{k=m}^{n} a_k + \sum_{k=m}^{n} b_k, \qquad \sum_{k=m}^{n} c \, a_k = c \sum_{k=m}^{n} a_k$$
Splitting the range: $\sum_{k=1}^{n} a_k = \sum_{k=1}^{m} a_k + \sum_{k=m+1}^{n} a_k$. This lets you compute a piece of a sum: $\sum_{k=m}^{n} a_k = \sum_{k=1}^{n} a_k - \sum_{k=1}^{m-1} a_k$.
Lower bound $k = 0$: split off the $k = 0$ term. Example: $\sum_{k=0}^{8}(k^2 - 1) = (0 - 1) + \sum_{k=1}^{8}(k^2 - 1)$.
Memorize these three formulas:
$$\sum_{k=1}^{n} k = \frac{n(n+1)}{2}, \qquad \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}, \qquad \sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2$$
Combined with linearity they cover most exam sums. Example: $\sum_{k=1}^{20}(2k+7) = 2 \cdot \frac{20 \cdot 21}{2} + 7 \cdot 20 = 420 + 140 = 560$.
A piece of a sum is a difference of two full sums. Example:
$$4^2 + 5^2 + \dots + 20^2 = \sum_{k=1}^{20} k^2 - \sum_{k=1}^{3} k^2 = \frac{20 \cdot 21 \cdot 41}{6} - \frac{3 \cdot 4 \cdot 7}{6} = 2870 - 14 = 2856$$
In an arithmetic progression each term differs from the previous one by a constant $d$ (the common difference):
$$a_n = a_1 + (n-1)d, \qquad S_n = \frac{(a_1 + a_n) \cdot n}{2}$$
Number of terms from $a_1$ to $a_n$ with step $d$: $n = \frac{a_n - a_1}{d} + 1$.
Example: $1 + 4 + 7 + \dots + 130$. Here $a_1 = 1$, $d = 3$, so $n = \frac{130 - 1}{3} + 1 = 44$ and $S = \frac{(1 + 130) \cdot 44}{2} = 131 \cdot 22 = 2882$.
Alternating sums are handled by pairing: $1 - 2 + 3 - 4 + \dots + 43 - 44 = (1-2) + (3-4) + \dots + (43-44) = 22 \cdot (-1) = -22$, since there are $44/2 = 22$ pairs, each equal to $-1$. If the last term is unpaired, split it off separately.
In a geometric progression each term is the previous one multiplied by a constant ratio $q \ne 1$:
$$a_n = a_1 q^{\,n-1}, \qquad S_n = a_1 \cdot \frac{q^n - 1}{q - 1}$$
The most common exam form is a sum of powers:
$$\sum_{k=0}^{n} q^k = 1 + q + q^2 + \dots + q^n = \frac{q^{\,n+1} - 1}{q - 1}$$
For $q = 2$ this simplifies nicely: $\sum_{k=0}^{n} 2^k = 2^{\,n+1} - 1$. Example: $\sum_{k=0}^{8} 2^k = 2^9 - 1 = 511$ (note: $9$ terms, from $2^0$ to $2^8$).
In a telescoping sum each term is written as a difference $f(k) - f(k+1)$; the inner parts cancel and only the first and last survive.
Key decomposition: $\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$, hence
$$\sum_{k=1}^{n} \frac{1}{k(k+1)} = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \dots + \left(\frac{1}{n} - \frac{1}{n+1}\right) = 1 - \frac{1}{n+1} = \frac{n}{n+1}$$
Logarithmic telescoping: since $1 + \frac{1}{k} = \frac{k+1}{k}$, we get $\sum_{k=1}^{n} \log_2\left(1 + \frac{1}{k}\right) = \log_2\left(\frac{2}{1} \cdot \frac{3}{2} \cdots \frac{n+1}{n}\right) = \log_2(n+1)$.
Telescoping product: the same fraction chain cancels: $\prod_{n=1}^{m}\left(1 + \frac{1}{n+5}\right) = \frac{7}{6} \cdot \frac{8}{7} \cdots \frac{m+6}{m+5} = \frac{m+6}{6}$.
The symbol $\prod_{k=m}^{n} a_k = a_m \cdot a_{m+1} \cdots a_n$ denotes a product; the number of factors is again $n - m + 1$. Properties: $\prod (a_k b_k) = \prod a_k \cdot \prod b_k$ and $\prod_{k=m}^{n} c = c^{\,n-m+1}$.
Zero factor — check first! If even one factor equals $0$, the whole product is $0$. Example: $\prod_{k=1}^{10} (2k^2 - 7k - 4)$. Factor: $2k^2 - 7k - 4 = (2k+1)(k-4)$, which vanishes at $k = 4$. Since $4$ lies in the range $1 \le k \le 10$, the product is $0$. Similarly $\prod \left[(k-24)(k-25)\right] = 0$ as soon as $k = 24$ or $k = 25$ is inside the range.
Sign of a product of $-1$'s: $\prod_{k=m}^{n} (-1) = (-1)^{\,n-m+1}$. Example: $\prod_{k=15}^{41} (-1) = (-1)^{27} = -1$, because $41 - 15 + 1 = 27$ is odd. In general, the sign of a product is determined by the parity of the number of negative factors.
In a nested (double) expression, evaluate from the inside out: first compute the inner sum as a function of the outer index, then run the outer operation.
Example: $\sum_{k=1}^{4} \sum_{m=1}^{k} 2m$. Inner sum: $\sum_{m=1}^{k} 2m = 2 \cdot \frac{k(k+1)}{2} = k(k+1) = k^2 + k$. Outer sum:
$$\sum_{k=1}^{4} (k^2 + k) = 2 + 6 + 12 + 20 = 40$$
The same strategy works when a product wraps a sum, e.g. $\prod_{n=1}^{a} \left(\sum_{k} a_k\right)$: simplify the inner sum to a formula in $n$, then look at the product — often it telescopes or contains a zero factor.
When a sum contains an unknown parameter or an unknown upper bound, expand the sum with the standard formulas and solve the resulting equation.
Unknown parameter. $\sum_{n=1}^{20} (2 + na) = 70$. Expand: $\sum_{n=1}^{20} 2 + a \sum_{n=1}^{20} n = 40 + a \cdot \frac{20 \cdot 21}{2} = 40 + 210a$. Then $40 + 210a = 70$, so $a = \frac{30}{210} = \frac{1}{7}$.
Unknown upper bound. $\sum_{k=7}^{n} (k+3) = 46$. The terms are $10, 11, 12, \dots$ For small totals just add: $10 + 11 + 12 + 13 = 46$ — four terms, so $n = 7 + 4 - 1 = 10$. For larger totals use $S = \frac{(a_1 + a_n) \cdot (n - m + 1)}{2}$ and solve the quadratic; keep only the natural root that is not less than the lower bound.
Problem. The sequence $(a_n) = \left(\frac{2n+1}{3n+4}\right)$ is given. If $a_k = \frac{1}{2}$, find $k$.
Solution.
Step 1. Substitute $n = k$ and set the general term equal to $\frac{1}{2}$: $\frac{2k+1}{3k+4} = \frac{1}{2}$.
Step 2. Cross-multiply: $2(2k+1) = 3k+4$, that is $4k + 2 = 3k + 4$.
Step 3. Solve: $k = 2$.
Step 4. Check: $a_2 = \frac{2 \cdot 2 + 1}{3 \cdot 2 + 4} = \frac{5}{10} = \frac{1}{2}$. Answer: $k = 2$.
Similarly, for $(a_n) = (3n+7)$ the next term is obtained by substituting $n+1$: $a_{n+1} = 3(n+1) + 7 = 3n + 10$.
Problem. Compute $\sum_{k=0}^{8} (k^2 - 1)$.
Step 1. Count the terms: from $k = 0$ to $k = 8$ there are $8 - 0 + 1 = 9$ terms.
Step 2. Use linearity: $\sum_{k=0}^{8} (k^2 - 1) = \sum_{k=0}^{8} k^2 - \sum_{k=0}^{8} 1 = \sum_{k=0}^{8} k^2 - 9$.
Step 3. The $k = 0$ term contributes $0$ to the square sum, so $\sum_{k=0}^{8} k^2 = \sum_{k=1}^{8} k^2 = \frac{8 \cdot 9 \cdot 17}{6} = 204$.
Step 4. Result: $204 - 9 = 195$. Answer: $195$.
By the same method: $\sum_{k=1}^{20}(2k+7) = 2 \cdot \frac{20 \cdot 21}{2} + 7 \cdot 20 = 560$.
Problem A. Compute $\sum_{k=1}^{15} \log_2\left(1 + \frac{1}{k}\right)$.
Step 1. Rewrite: $1 + \frac{1}{k} = \frac{k+1}{k}$, and a sum of logarithms is the logarithm of the product.
Step 2. The product telescopes: $\frac{2}{1} \cdot \frac{3}{2} \cdot \frac{4}{3} \cdots \frac{16}{15} = 16$.
Step 3. $\log_2 16 = 4$. Answer: $4$.
Problem B. Compute $\prod_{k=15}^{41} (-1)$.
Step 1. The number of factors is $41 - 15 + 1 = 27$.
Step 2. $(-1)^{27} = -1$, since $27$ is odd. Answer: $-1$.
Problem A. If $\sum_{n=1}^{20} (2 + na) = 70$, find $a$.
Step 1. Split by linearity: $\sum_{n=1}^{20} 2 + a \sum_{n=1}^{20} n$.
Step 2. Compute each part: $\sum_{n=1}^{20} 2 = 40$ and $\sum_{n=1}^{20} n = \frac{20 \cdot 21}{2} = 210$.
Step 3. Solve $40 + 210a = 70$: $210a = 30$, so $a = \frac{1}{7}$. Answer: $a = \frac{1}{7}$.
Problem B. Compute $\sum_{k=1}^{4} \sum_{m=1}^{k} 2m$.
Step 1. Inner sum: $\sum_{m=1}^{k} 2m = k(k+1)$.
Step 2. Outer sum: $\sum_{k=1}^{4} k(k+1) = 1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + 4 \cdot 5 = 2 + 6 + 12 + 20 = 40$. Answer: $40$.
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