Solving $\left(1+3+5...+(2n+1)\right ) + \left(3.5+5+6.5+...+(\frac{7+3n}{2})\right)=105$

Question

$\left(1+3+5...+(2n+1)\right ) + \left(3.5+5+6.5+...+(\frac{7+3n}{2})\right)=105$

It is the equation that I did not understand how to find $n.$

Answer 1

It's $$(n+1)^2+\frac{(2\cdot3.5+n\cdot1.5)(n+1)}{2}=105$$ I got $n=6$.

I used the following formula.

$$S_n=\frac{(2a_1+(n-1)d)n}{2}.$$ Thus, $$1+3+...+(2n+1)=\frac{(2\cdot1+n\cdot2)(n+1)}{2}=(n+1)^2.$$ Also, we can use the following nice reasoning. $$\sum_{k=1}^{n+1}(2k-1)=\sum_{k=1}^{n+1}(k^2-(k-1)^2)=(n+1)^2.$$

Answer 2

HINT: use that $$1+3+5+...+2n+1=(n+1)^2$$ and $$3.5+5+...+\frac{7+3n}{2}=\frac{1}{4}(3n+14)(n+1)$$

Answer 3

Using the fact that $\sum_{k=0}^n k =\tfrac12 n(n+1)$ \begin{align}\sum_{k=0}^n(2k+1) + \sum_{k=0}^n \frac{7+3k}{2} &= \left(2\sum_{k=0}^n k + (n+1)\right) + \left(\frac72 (n+1) + \frac32\sum_{k=0}^nk\right)\\ &= \left(n(n+1) + (n+1)\right) + \left(\frac72 (n+1) + \frac34 n(n+1)\right)\\ &= (n+1)^2 + \frac14 (3n+14)(n+1)\\ &=\frac14 (n+1)(7n+18)\\ &= \frac{7n^2+25n+18}{4} \end{align} Then solve the quadratic equation $\frac{7n^2+25n+18}{4} = 105$: $$7n^2+25n+18 = 420 \implies 7n^2+25n-402 = 0$$

which has roots at $n=-\frac{67}{7}$ and $6$. Clearly $n$ must be positive, so $n=6$.

Answer 4

If you double both sides and pair up terms from the two sums you're looking at the equation

$$210=(2+7)+(6+10)+(10+13)+(14+16)+\cdots+((4n+2)+(7+3n))\\=9+16+23+30+37+44+\cdots+(7n+9)$$

so you can just track the running sum until you get to $210$, i.e.,

$$9+16=25, 25+23=48, 48+30=78, 78+37=115, 115+44=159, 159+51=210$$

This tells you to stop at $51=7\cdot6+9$, so the solution is $n=6$.

Remarks: What makes this work is that $210$ is small enough that it only takes a few additions to get the answer. If the problem had, say a six-digit number on the right hand side instead of $105$, you'd be better off using the summation-formulae approach of the other answers. Note also that the number $105$ was chosen so that there actually is a solution; if you had some random integer, the summation-formulae approach would give an answer for $n$, but it wouldn't be an integer.