Sum $\sum_{i=k}^n 3$

$\sum_{i=k}^n 3$ I calculated it, using the arithmetic sum formula to be $\frac{n(2*3+(n-1)*0)}{2} = 3n$ but the answer says its $3(n-k+1)$. Why?

Solution 1:

The formula you're using is $\frac n2 (2a + (n-1)d)$, but the "$n$" in the formula is not the same thing as the "$n$" in the problem. For clarity, let's rename all the variables in the formula to capital letters, so that we don't confuse them with the variables in the problem. The formula for the sum is $\frac N2(2A + (N-1)D)$, where

• $N$ is the number of terms;
• $A$ is the first term;
• $D$ is the common difference.

You are right that $D$, the common difference, should be $0$. You are also right that $A$, the first term, should be $3$. However, $N$, the number of terms, should not be $n$. Instead, since $i$ goes from $k$ to $n$, there are $n-(k-1)$ or $n-k+1$ terms: there are $n$ numbers from $1$ to $n$, but we skip the $k-1$ numbers from $1$ to $k-1$, starting at $k$.

So the formula tells us that the sum is $$ \frac N2 (2A + (N-1)D) = \frac{n-k+1}{2} (2\cdot 3 + (n-k+1)\cdot 0) = \frac{n-k+1}{2} \cdot 6 $$ which simplifies to the $3(n-k+1)$ you want to get.