$$ \sum_{i=1}^n (ai +b) $$
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
Best Answer
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 $n \cdot b$.