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