$$ \sum_{i=1}^n (ai +b) $$

Let $n \geq 1$ be an integer, and let $a,b > 0$ be positive real numbers. Find a closed form for the following expression. In other words you are to eliminate the summation and write the expression as a function of a b and n. Explain how you computed your answer

I have tried to complete the exercise, but am not receiving the proper answer. Can anyone please provide English based understanding? I am trying to parse it out as much as possible for better understanding

first term = 1 Last term = an + b constant difference = b

$$ i = (n(1 + (an + b))b) /2 $$

Therefore the above summation =

$$ a * ((n(1+(an+b))b)/2) + b =$$ $$ a(n(1+an+b)b)/2+b = $$ $$ a((n+an^2 +bn)b)/2+b = $$

$$ a(bn +abn^2 +b^2n)/2 + b $$

This appears to be failing. when a = 1 b=1 and n =2, summation should = 3, the formula above gives 5

What am i missing here? everything seems right I feel like I am missing something simple.

Also, can someone provide help on proof by induction, in as simple terms as possible?

Thanks in advance


The constant difference between each term is not $b$, and the first term is not $1$ (but rather $a+b$).

For clarity, I would recommend doing the following. Split the sum into two parts:

$$\sum_{i=1}^n ai + \sum_{i=1}^n b$$

This is equivalent to:

$$[a + 2a + 3a + \dots + (n-1)a + na] + [b + b + b + \dots + b +b]$$

The first sum is an arithmetic progression (A.P) where the difference between each term is $a$, and we have $n$ terms. You seem to know how to do this? The second sum is just the summing of $n$ terms of $b$, which is $n \cdot b$.

  • 1
    $\begingroup$ Hmm, wouldn't it be clearer to write $ \displaystyle a \cdot (\sum_{i=1}^n i) $ and $\displaystyle b \cdot (\sum_{i=1}^n 1)$ ? The closed-forms for the sum-expressions should even easier jump out from the memory... $\endgroup$ – Gottfried Helms Apr 26 '15 at 20:04
  • $\begingroup$ s the following correct now? Using the split summations above for clarity if we factor out a, first term = 1, last term = n, difference = 1.. $$a((n(1+n)1)/2)+nb $$ $$a((n+n^2)/2)+nb$$ $\endgroup$ – Busturdust Apr 26 '15 at 20:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.