Sequences and Notation: Additional Practice (Set 1)

Substitute the index $n = 7$ and simplify the numerator before dividing.

$a_7 = \frac{7(7+1)}{2} = \frac{7 \cdot 8}{2} = \frac{56}{2} = 28$

These are the triangular numbers $1, 3, 6, 10, 15, 21, 28, \ldots$, so $28$ is indeed the seventh. Forgetting to halve gives $56$, and $21$ is $a_6$.

Square the previous term, then subtract $1$, and repeat. There is no way to skip ahead.

$a_2 = 2^2 - 1 = 3, \qquad a_3 = 3^2 - 1 = 8$

One more rung gives the answer.

$a_4 = 8^2 - 1 = 64 - 1 = 63$

Stopping at $a_3 = 8$ is the usual off-by-one.

Split the summand first, since $\frac{1}{k} - \frac{1}{k+1} = \frac{(k+1) - k}{k(k+1)} = \frac{1}{k(k+1)}$.

$\left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{5}\right)$

Every interior piece cancels against its neighbor, leaving only the two ends.

$1 - \frac{1}{5} = \frac{4}{5}$

Adding directly confirms it: $\frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} = \frac{4}{5}$.

The rule is a perfect square, which makes the zero easy to spot.

$a_n = n^2 - 4n + 4 = (n-2)^2$

A square is zero exactly when the quantity inside it is zero, so $n - 2 = 0$ and $n = 2$. Checking the neighbors, $a_1 = 1$ and $a_3 = 1$, so no other index works.

Compare the two sequences index by index, since only positive integers are legal inputs.

$n = 1: \ 1 \text{ vs } 1 \qquad n = 2: \ 4 \text{ vs } 4 \qquad n = 3: \ 9 \text{ vs } 7$

At $n = 1$ and $n = 2$ the two agree, so neither is strictly greater. At $n = 3$ the squares pull ahead, $9 > 7$, so the smallest such index is $3$.

Evaluate the whole summand at each index, keeping the negative first term.

$(1 - 2) + (4 - 4) + (9 - 6) + (16 - 8) + (25 - 10)$

That is $-1 + 0 + 3 + 8 + 15$.

$-1 + 0 + 3 + 8 + 15 = 25$

A quick check: the squares sum to $55$ and $2k$ sums to $30$, and $55 - 30 = 25$.

The amount added changes at every step, because the rule adds the current index rather than a fixed number.

$a_2 = 5 + 2 = 7, \qquad a_3 = 7 + 3 = 10, \qquad a_4 = 10 + 4 = 14$

One more rung, adding $5$ this time.

$a_5 = 14 + 5 = 19$

Stopping at $a_4$ leaves you one rung short, at $14$. The other two traps come from adding the wrong indices. The rule first fires at $n = 2$, so the numbers added are $2, 3, 4, 5$; adding $1, 2, 3, 4$ instead gives $5 + 10 = 15$, and adding $1, 2, 3, 4, 5$ (one application too many) gives $5 + 15 = 20$.

Two definitions agree only if they produce the same term at every index, so generate a few terms from each side.

$a_n = n^2: \ 1, 4, 9, 16 \qquad \text{recursion}: \ 1,\ 1+3 = 4,\ 4+5 = 9,\ 9+7 = 16$

They match, and the algebra confirms it, since $n^2 - (n-1)^2 = 2n - 1$. The others break down by the third term: $2n$ gives $2, 4, 6$ while doubling gives $2, 4, 8$; $n+1$ starts at $2$ while the recursion starts at $1$; and $3n$ gives $3, 6, 9$ while the last recursion gives $3, 5, 8$.

Count the indices from the lower limit to the upper limit, with both ends included.

$40 - 12 + 1 = 29$

The summand $\frac{1}{k}$ has nothing to do with the count. Subtracting the limits alone gives $28$, which counts the steps rather than the terms.

Factor the numerator as a difference of squares and cancel, which is legal because $n + 1 \ne 0$ for any positive index.

$a_n = \frac{(n-1)(n+1)}{n+1} = n - 1$

So the tenth term is

$a_{10} = 10 - 1 = 9.$

Substituting directly agrees: $\frac{100 - 1}{11} = \frac{99}{11} = 9$.

Compute the four terms first, letting $(-1)^n$ set each sign.

$a_1 = -1, \qquad a_2 = 3, \qquad a_3 = -5, \qquad a_4 = 7$

Now add them, which is exactly what the sigma instructs.

$-1 + 3 - 5 + 7 = 4$

Ignoring the sign factor would give $1 + 3 + 5 + 7 = 16$.

Set the two rules equal and solve the resulting quadratic.

$n^2 - 5n = 6 \;\Longrightarrow\; n^2 - 5n - 6 = 0 \;\Longrightarrow\; (n - 6)(n + 1) = 0$

The roots are $n = 6$ and $n = -1$, but $-1$ is not a positive integer, so it is not an index at all and has to be discarded. Checking the survivor, $a_6 = 36 - 30 = 6$, which matches $b_6 = 6$.