Math Induction Series Proof

Question

$$\sum_{i = 1}^n (2i+3) = n(n+4)$$ for all n >= 1.

Was a homework problem that was given no solution. Was told last lines weren't correctly written. My attempt: Let P(n) = n(n+4) for all n >= 1 Basis Step: P(2) = 2(6) = 12 >= 1 Inductive Step: $$\sum_{(i=1}^{k+1} (k+1)(k+5)$$

= k(k+4) + (k+1) = $$k^2 + 4k + (k+1) $$ = $$k^2 + 5k+ 1 $$ :(k+1)(k+5) QED >= 1

Edit: The question is what would be the correct steps to solve it or fix what I have above so I know for my exam Friday.

Answer 1

First, show that this is true for $n=1$:

$\sum\limits_{i=1}^{1}2i+3=1(1+4)$

Second, assume that this is true for $n$:

$\sum\limits_{i=1}^{n}2i+3=n(n+4)$

Third, prove that this is true for $n+1$:

$\sum\limits_{i=1}^{n+1}2i+3=$

$\color\red{\sum\limits_{i=1}^{n}2i+3}+2(n+1)+3=$

$\color\red{n(n+4)}+2(n+1)+3=$

$n^2+6n+5=$

$(n+1)(n+5)=$

$(n+1)(n+1+4)$

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Please note that the assumption is used only in the part marked red.

Answer 2

The inductive step is $$\sum_{i=1}^n(2i+3)=n(n+4)$$ and you need to show that $$\sum_{i=1}^{n+1}(2i+3)=(n+1)(n+5).$$ We have

$$\sum_{i=1}^{n+1}(2i+3)=\color{red}{\sum_{i=1}^{n}(2i+3)}+2n+5.$$ Use the induction step to susbtitute the value of the sum in red and you are done.

Answer 3

Assume the statement holds for $n=k$:

$$\sum_{i = 1}^k (2i+3) = k(k+4).$$

Write out the left side for $n=k+1$ and split off the last term:

$$\sum_{i = 1}^{k+1} (2i+3) = \left[\sum_{i = 1}^{k} (2i+3)\right] + 2(k+1) + 3.$$

The manipulation above is just algebra, but we now have a piece that we can substitute in the assumption directly.

Now use the inductive assumption to substitute the bracketed sum on the right side:

$$\sum_{i = 1}^{k+1} (2i+3) = k(k+4) + 2(k+1) + 3 \\= k^2 + 6k+5 \\= (k+1)(k+5) \\= (k+1)[(k+1)+4].$$

Done!

Answer 4

Let $n\geq1$ such that

$$S_n=\sum_{k=1}^n(2i+3)=n(n+4). $$ then

$$S_{n+1}=(n+1)(n+5)$$

$$=n(n+4)+2n+5$$

$$=n(n+4)+2(n+1)+3$$

$$=S_n+2(n+1)+3$$

$$=\sum_{k=1}^{n+1}(2i+3)$$