Sequences and Notation: Practice

Write out the terms the original produces, since that is what has to be matched.

$\sum_{k=1}^{5} (k+2) = 3 + 4 + 5 + 6 + 7 = 25$

Adding $2$ inside the summand is the same as shifting both limits up by $2$ and dropping the $+2$, which is exactly $\sum_{j=3}^{7} j$. The others give $15$, $35$, and $20$.

The summand $2k - 1$ produces the odd numbers $1, 3, 5, 7$, and each one is then squared.

$1^2 + 3^2 + 5^2 + 7^2 = 1 + 9 + 25 + 49 = 84$

The trap is the $-1$ sitting inside the parentheses. Drop it and you are summing $(2k)^2$ instead, which gives $4 + 16 + 36 + 64 = 120$. The whole parenthesis has to be squared, one term at a time.

The rule adds the next odd number each time, so climb the ladder from the seed.

$a_2 = 1 + 3 = 4, \qquad a_3 = 4 + 5 = 9, \qquad a_4 = 9 + 7 = 16$

Continuing to the sixth term,

$a_5 = 16 + 9 = 25, \qquad a_6 = 25 + 11 = 36.$

The terms are the perfect squares, so the explicit rule here is $a_n = n^2$. Compare that with the lesson, which ran this same step rule from the seed $a_1 = 2$ and landed on $a_n = n^2 + 1$. Same step rule, different seed, and every single term shifts by one. The seed is not decoration: it is half the definition.

Generate the terms first, then test the candidates against them.

$a_1 = 2, \quad a_2 = 5, \quad a_3 = 8, \quad a_4 = 11$

The rule $a_n = 3n - 1$ gives $2, 5, 8, 11$, so it matches. It also obeys the same step rule, since $(3n - 1) - \big(3(n-1) - 1\big) = 3$, and it starts at the same seed, so the two definitions agree at every index. The rule $3n + 2$ starts at $5$, the rule $2n + 1$ starts at $3$, and $3n$ starts at $3$.

The summand ignores the index, so the sum is simply $n$ copies of $5$ added together.

$\sum_{k=1}^{n} 5 = 5n$

Set that equal to $60$ and solve.

$5n = 60 \;\Longrightarrow\; n = 12$

At $k = 1$ the exponent $k + 1 = 2$ is even, so the first sign is positive, and the signs then alternate.

$1 - 2 + 3 - 4 + 5 - 6$

Pairing the terms makes the arithmetic quick: each pair contributes $-1$.

$(1 - 2) + (3 - 4) + (5 - 6) = -1 - 1 - 1 = -3$

Set the rule equal to $1$ and clear the denominator, which is safe because $n + 3 \ne 0$ for a positive index.

$\frac{2n}{n+3} = 1 \;\Longrightarrow\; 2n = n + 3 \;\Longrightarrow\; n = 3$

Check it: $a_3 = \frac{6}{6} = 1$. The other indices give $\frac{1}{2}$, $\frac{4}{3}$, and $\frac{3}{2}$.

As $k$ runs from $0$ to $4$, the quantity $k + 1$ runs from $1$ to $5$, so the terms being squared are $1, 2, 3, 4, 5$.

$\sum_{k=0}^{4} (k+1)^2 = 1 + 4 + 9 + 16 + 25 = 55$

That is exactly $\sum_{j=1}^{5} j^2$: substituting $j = k + 1$ shifts the limits up by one and clears the $+1$ out of the summand. The others give $30$, $90$, and $225$.

The summand factors as $(k+1)^2$, so the terms are the squares of $2$, $3$, and $4$.

$\sum_{k=1}^{3} (k+1)^2 = 4 + 9 + 16 = 29$

Evaluating term by term without factoring gives the same thing: $(1 + 2 + 1) + (4 + 4 + 1) + (9 + 6 + 1) = 4 + 9 + 16 = 29$.

A recursion can be run backwards as easily as forwards. Apply the rule at $n = 4$ first.

$a_4 = a_3 + a_2 \;\Longrightarrow\; 11 = 7 + a_2 \;\Longrightarrow\; a_2 = 4$

Now apply it at $n = 3$.

$a_3 = a_2 + a_1 \;\Longrightarrow\; 7 = 4 + a_1 \;\Longrightarrow\; a_1 = 3$

The sequence is $3,\ 4,\ 7,\ 11, \ldots$, which checks out.

Both sums run over the same indices, so their terms can be paired off and added index by index.

$(2 + 4 + 6 + 8 + 10) + (3 + 3 + 3 + 3 + 3) = 30 + 15 = 45$

Pairing the $k$th terms gives $(2k) + 3$ at each index, so the combined sum is $\sum_{k=1}^{5} (2k+3) = 5 + 7 + 9 + 11 + 13 = 45$. The others give $75$ (from wrongly merging $2k$ and $3$ into $5k$), $105$ (from dropping the whole second sum, $15$, into every term instead of adding it once), and $60$.

Apply the rule one rung at a time, simplifying each complex fraction before moving on.

$a_2 = \frac{1}{1 + 1} = \frac{1}{2}, \qquad a_3 = \frac{1/2}{1 + 1/2} = \frac{1/2}{3/2} = \frac{1}{3}$

One more step gives the term asked for.

$a_4 = \frac{1/3}{1 + 1/3} = \frac{1/3}{4/3} = \frac{1}{4}$

The terms are $\frac{1}{n}$, but the recursion alone never tells you that; you have to climb to find out.