Evaluating $\sum_{n=0}^{15}(1.6^n-12n+1)$. What am I doing wrong?

Question

Find the value of $$\sum_{n=0}^{15}(1.6^n-12n+1)$$ giving your answer correct to one decimal place.

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I separated the summation into two parts:

The summation of $1.6^n$ and the summation of $1-12n$.

I then put the equations into the summation formula:

I got $$\frac{1-1.6^{16}}{1-1.6} + \frac{16}{2}(1+(15)(-12))$$

This gave me an answer of $1640.8$, but the given answer is $1648.8$.

Where did I go wrong?

Answer 1

The $1$ shouldn't have its contribution halved, so the correct answer is $8$ more than you said.

Answer 2

How/why did you figure $\sum_{n=0}^{15}(1-12n)=\frac {16}2 (1+15(-12))$?

I don't like memorizing formulas so the only one I memorize is $\sum_{n=0|1}^M n = \frac {M(M+1)}2$. And that way $\sum_{n=0}^{15}(1-12n) = \sum_{n=0}^{15}1 - 12\sum_{n=0}^{15}n= 16 - 12(\frac {15\cdot 16}2)$.

But some people memorize (where I would derive) $\sum_{n=0}^M (a+bn) = \sum_{n=a}^{a+M}bn = b(M+1)\frac {(a + 0\cdot b)+(a+Mn)}2= \frac {M+1}2b(2a+Mn)=\frac {16}2(2+15(-12))$.

So I guess you mangled the formula. We "average" the changes in the value so the $-12*0$ to $-12*15$ are added and divided in half. But the constant offset of $1$ stays canstant and there are $16$ of the so ..... we don't divide those by half.

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Okay, a lot of text have a formula for the sum arithemetic series given as

$\sum_{k=0}^{n-1} (a+dk) = $(half the number of terms)$\times$(the first term plus the last term$=$

$\frac n2([a + 0*d] + [a + (n-1)*d])=$

$\frac n2(\color{red} 2a + (n-1)*d)$.

You are not taking the $\color{red}2$ into account. The $a$ term contributes to the first term AND to the last term. So you must take $a$ into account twice.

....

But.... don't just memorize formulas. Always reason them out.

$\sum_{k=0}^{15} (1-12n) = $(number of terms)$\times$(average number of terms) and as this is an arithmetic progression $=$

$16 \frac {\text{first term} + \text{last term}}2=$

$8(1 + [1+(-12)15])=$

$8 (2 -12*15)$.