How to find the largest value of $n$ such that $S_n = \sum_{k = 1}^n (2k + 1) < 68$

Question

Please help me solve the following:

Given the following Sigma Notation:

Given the following: $$\sum_{k=1}^n (2k+1)$$

what is the greatest number of terms for which $S_n<68$?

I tried the following:

$\sum_{k=1}^n (2k+1)$

$S_1=2(1)+1=3$

$S_2=2(2)+1=5$

$S_3=2(3)+1=7$

$t_1+t_2+t_3+ \ldots +2(k)+1= 3+5+7+...2(k)+1$

$a=3$

$d=2$

$S_n=\frac{n(2a+(n-1)d}{2}$

$68>\frac{n(2(3)+2n-2}{2}$

$34>n(4+2n)$

$17>n(2+n)$

$17>2n+n^2$

This where I get stuck, I feel like my approach is completely wrong. Please assist.

Answer 1

We are given $$S_n = \sum_{k = 1}^{n} (2k + 1)$$ Since we want to find the largest value of $n$ such that $S_n < 68$, $$S_n = \sum_{k = 1}^{n} (2k + 1) < 68$$ Since the sequence $(a_k)$ defined by $a_k = 2k + 1$ is arithmetic, this is an arithmetic series. Hence, the $n$th partial sum is $$S_n = \frac{n(a_1 + a_n)}{2} = \frac{n[a_1 + a_1 + (n - 1)d]}{2} = \frac{n[2a_1 + (n - 1)d]}{2}$$ Observe that $a_1 = 2 \cdot 1 + 1 = 3$ and $a_n = 2n + 1$. Hence, \begin{align*} S_n & < 68\\ \frac{n(a_1 + a_n)}{2} & < 68\\ \frac{n(3 + 2n + 1)}{2} & < 68\\ \frac{n(2n + 4)}{2} & < 68\\ n(n + 2) & < 68\\ n^2 + 2n & < 68\\ n^2 + 2n + 1 & < 68 + 1 && \text{complete the square}\\ (n + 1)^2 & < 69 \end{align*} The largest integer with square less than $69$ is $8$. Thus, $n + 1 = 8 \implies n = 7$.

Check: By direct calculation, $$\sum_{k = 1}^7 (2k + 1) = 3 + 5 + 7 + 9 + 11 + 13 + 15 = 61 < 68$$ while $$\sum_{k = 1}^8 (2k + 1) = 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 = 78 > 68$$ Thus, $n = 7$ is correct.

Where did you make a mistake?

After using the formula $$S_n = \frac{n[2a_1 + (n - 1)d}{2}$$ to obtain $$S_n = \frac{n(2 \cdot 3 + (n - 1) \cdot 2}{2}$$ you obtained $$S_n = \frac{n(6 + 2n - 2)}{2}$$ Therefore, you obtained $$\frac{n(2 \cdot 3 + 2n - 2)}{2} < 68$$ You then multiplied the expression on the left-hand side in my work by $2$ while dividing $68$ by $2$, so you were off by a factor of $4$.

You should have obtained \begin{align*} \frac{n(6 + 2n - 2)}{2} & < 68\\ \frac{n(2n + 4)}{2} & < 68\\ n(n + 2) & < 68 \end{align*} and then proceeded as above.

Answer 2

Just to add an observation to the answer of N. F. Taussig: consider a square, it can be divided in the following way

and this proves that the sum of the first $n$ odd numbers (starting from $1$) is $n^2$. Your sum is the sum of the first $n+1$ odd numbers where you are neglecting the first (you have $n$ addends from $1$ to $n$), so it is $(n+1)^2-1$ and you have to solve $(n+1)^2<69$ in integers.