Summation problem: find k if $\sum_{k=5}^{29} kn-6=1125$

Question

How do I find k if $\sum_{k=5}^{29} kn-6=1125$ ? I tried to solve it but couldn’t understand it. Any hints would be appreciated!

Answer 1

If I understood the question right: $$\sum_{k=5}^{29}(kn-6)=1125$$ $$\iff n\cdot\sum_{k=5}^{29}(k)-6\cdot(29-4)=1125$$ $$\iff n\cdot\sum_{k=5}^{29}(k)=1275$$ $$\iff n\left(\frac{29(29+1)}{2}-\frac{4(4+1)}{2}\right)=1275 \tag{sum from 1 to $n$ formula}$$ $$\iff n=3 $$ What I did here in deep details: $$(5n-6)+(6n-6)+\dots+(29n-6)=1125$$ $$\iff 5n+6n+\dots+29n-(6+6+\dots+6)=1125 \tag{rearragning terms}$$ Now, how many $6$s are there? Well the first $6$ is at $k=5$ and the last is at $k=29$, so we need to make $k=1$ so that we can count, and we can do that by subtracting $4$ to get that it's from $k=1$ to $k=25$ for identical $6$s: $$\iff 5n+6n+\dots+29n-6\cdot25=1125 $$ $$\iff 5n+6n+\dots+29n=1125+6\cdot25=1275 $$ $$\iff n(5+6+\dots+29)=1275 $$ $$\iff n((1+2+3+4)+5+6+\dots+29-(1+2+3+4))=1125+6\cdot25=1275 $$ $$\iff n(\ \underbrace{(1+2+3+4+5+6+\dots+29)}_\text{formula for sum from 1 to $n$}-\underbrace{(1+2+3+4)}_\text{same formula})=1275 $$ $$\iff n\left(\frac{29(29+1)}{2}-\frac{4(4+1)}{2}\right)=1275 $$ $$\iff n=3$$ So, the thing is, you have to learn some properties of summation in order to be able to solve these problems easily.

If the required is $$\sum_{k=5}^{29}(kn)-6=1125$$ Then we'd do the following: $$\sum_{k=5}^{29}(kn)=1131$$ $$\iff n \cdot \sum_{k=5}^{29}(k)=1131$$ $$\iff n \cdot \left(\frac{29(29+1)}{2}-\frac{4(4+1)}{2}\right) =1131$$ $$\iff n =\frac{1131}{425}$$

Answer 2

$$\sum_{k=5}^{29}(nk-6)=\left[n\left(\sum_{k=1}^{29}k\right)-29\cdot6\right]-\left[n\left(\sum_{k=1}^{4}k\right)-4\cdot 6\right]$$

This gives

$$n\left[\frac{29(30)}{2}\right]-174-n\left[\frac{(4)5}{2}\right]+24=1125$$

$$435n-174-10n+24=1125$$

$$425n=1275$$

$$n=3$$

Answer 3

One question about the question is whether you mean $\left(\sum_{k=5}^{29} kn\right)-6=1125$ or $\sum_{k=5}^{29} (kn-6)=1125$. The other is that there is no $k$ to find because $k$ is a dummy variable of the summation. It ranges from $5$ to $29$. More reasonable is to try to find $n$. That is a free variable in the equation. As you have one equation in one unknown, you may well be able to find it.

Let us assume the second. You should be able to extend this to the first if that is your question. We can distribute out the $n$ getting $$n\sum_{k=5}^{29} k-\sum_{k=5}^{29} 6=1125$$ The sum of the $6$s is just that there are $25$ terms, so it is $150$. If you know that $\sum_{k=1}^{m} k=\frac 12m(m+1)$ you can write $\sum_{k=5}^{29} k=\sum_{k=1}^{29} k-\sum_{k=1}^{4} k=\frac 12\cdot 29\cdot 30-\frac 12 \cdot 4 \cdot 5=425$ This gives $$425n-150=1125\\n=3$$ The fact that it comes out even makes me suspect the $6$ was supposed to be in the sum.