Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Use the recursive definition of summation together with mathematical induction to prove that for all positive integers $n$ if $a_1, a_2,\ldots, a_n$ are real numbers, then

$$\sum_{k=1}^n(3a_k - 2k + 3) = 3\sum_{k=1}^na_k - n^2 + 2n.$$

Attempted Solution: I know how to check the base case ($n = 1$) of both the left and the right side to get $3a_1 + 1$.

For my inductive hypothesis, I got:

Assume for $j \ge1$, $\sum_{k=1}^j(3a_k - 2k + 3) = 3\sum_{k=1}^ja_k - j^2 + 2j$.

From the recursive definition of summation: $S_{j+1}= S_j + a_{j + 1}$.

After this I am lost. I don't know where to start, and unfortunately, my professor didn't have any notes about proving these with mathematical induction. Looking at the solutions manual I am even more confused. If anyone can look at his solution and kind of go step by step with me to tell me what he did or where he got things from, I would seriously appreciate it. I am especially having trouble knowing with what equations to set things up with, and how to change the $j$ on sigma to $j + 1$.

Teachers Solution:

$$\begin{align} S_{j+1} &= S_j + a_{j+1}\\ &=\sum_{k=1}^j(3a_k - 2k + 3) + 3a_{j+1}-2(j+1)+3\\ &= 3\sum_{k=1}^ja_k - j^2 + 2j + 3a_{j+1}-2(j+1)+3\\ &= 3\sum_{k=1}^{j+1}a_k - j^2 + 2j -2(j+1)+3\\ &= 3\sum_{k=1}^{j+1}a_k - (j+1)^2 + 2(j+1)\\ &=\text{Right Side}_{j+1} \end{align}$$

Therefore, $\sum_{k=1}^n(3a_k - 2k + 3) = 3\sum_{k=1}^na_k - n^2 + 2n.$.

Thank you for any help..

share|improve this question
add comment

2 Answers 2

up vote 5 down vote accepted

The teacher's solution is somewhat problematic in itself as it (ab)uses $a_j$ for two purposes: Once for the "generic" name of the summand in the first line $S_{j+1}=S_j+a_{j+1}$ and then as the numbers from which the summands are computed via $3a_k-2k+3$. Of course in general $a_{j+1}\ne3a_{j+1}-2(j+1)+3$!

I guess the most promising way to prove equations about such summations is as follows: Consider the left hand side and the right hand side as functions of $n$, say $f(n)=\sum_{k=1}^n(3a_k-2k+3)$ and $g(n)=3\sum_{k=1}^na_k-n^2+2n$. Then show that $f(1)=g(1)$ (you did that). Next concentrate on one side and compute $f(n)-f(n-1)$ for $n>1$, making use of the recursive definition: $$ \begin{align}f(n)-f(n-1)&=\sum_{k=1}^{n}(3a_k-2k+3)-\sum_{k=1}^{n-1}(3a_k-2k+3)\\&= \left(\sum_{k=1}^{n}(3a_k-2k+3)+(3a_n-2n+3)\right)-\sum_{k=1}^{n-1}(3a_k-2k+3)\\&=3a_n-2n+3.\end{align}$$ Do the same with the right hand side: $$\begin{align}g(n)-g(n-1)&=\left(3\sum_{k=1}^na_k-n^2+2n\right)-\left(3\sum_{k=1}^{n-1}a_k-(n-1)^2+2(n-1)\right) \\ &=\left(3\sum_{k=1}^{n-1}a_k+3a_n-n^2+2n\right)-\left(3\sum_{k=1}^{n-1}a_k-(n-1)^2+2(n-1)\right) \end{align}$$ and simplify until you notice that this is just $f(n)-f(n-1)$.

share|improve this answer
    
Since Fn+1 = Fn + Fn-1 by definition, why are we not using f(n) + f(n-1)? Is there a definition that allows us to use f(n) - f(n-1)? –  Alex May 3 at 10:01
add comment

Only a hint to make things easyer:

$$\sum_{k=1}^{n}(3a_{k}-2k+3)=3\sum_{k=1}^{n}a_{k}-n^{2}+2n$$ is clearly a direct consequence of: $$\sum_{k=1}^{n}\left(-2k+3\right)=-n^{2}+2n$$

Simply because: $$\sum_{k=1}^{n}3a_{k}=3\sum_{k=1}^{n}a_{k}$$ and this term can be added or subtracted on both sides.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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