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

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

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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 $n \cdot b$.

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    $\begingroup$ 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

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