Finding anti difference for a given expression

Question

I am given:

$$ a_n = \frac{1}{n^2 +3n+2}$$

And am asked to find the anti difference (i.e: quantity which when differences over gives the expression)

So, I started by partial fractions

$$ a_n = \frac{1}{n+1} + \frac{-1}{n+2} $$

Now the answer to this is:

$$ -\Delta \frac{1}{n+1}$$

But its wrong for some reason ( doesn't match with given answers)..

Options:

What difference is being used? Forward difference, backward difference, ...? What are the options? There's no way to specify what reason this answer is interpreted as wrong unless you clarify (telling us it doesn't match the given answers doesn't actually give any information). — Simply Beautiful Art, Nov 03 '20 at 14:09
Is the difference operator defined as a forward or backward difference? — Mark Viola, Nov 03 '20 at 14:10
what about the first answer $1-\frac 1{n+1}$ ? (the $1$ get cancelde in difference). — zwim, Nov 03 '20 at 14:56

score 1 · Accepted Answer · answered Nov 03 '20 at 14:11

1

If you denote $$b_n = -\frac{1}{n+1}$$

then $$b_{n+1}-b_n = a_n = \frac{1}{n^2 +3n+2}$$

So $b_n$ is the desired antidifference of $a_n$.

answered Nov 03 '20 at 14:11

peter.petrov

score 1 · Answer 2 · answered Nov 03 '20 at 14:19

1

Hint: If $b_{n+1}-b_n = a_n$ then $b_n = b_0 + a_0 + \cdots + a_{n-1}$. Since $a_n = \frac{1}{n+1} + \frac{-1}{n+2}$, the sum telescopes.

answered Nov 03 '20 at 14:19

lhf

2 Answers2