Find a pattern and prove it by mathematical induction:

Question

$1 = 1$

$3 + 5 = 8$

$7 + 9 + 11 = 27$

$13 + 15 + 17 + 19 = 64$

Etc...

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I am having trouble seeing a pattern with this, I know it is relatable with Fibonacci Numbers but I am having trouble grasping this topic

Answer 1

The first term it's $n^2-n+1$

and we have a sum of $n$ terms of an arithmetic progression with $d=2$.

Thus, $$(n^2-n+1)+(n^2-n+1+2)+...+(n^2-n+1+2(n-1))=$$ $$=n(n^2-n+1)+2\cdot\frac{n(n-1)}{2}=n^3.$$

Answer 2

The pattern is \begin{align} & (n^2-n+1) + (n^2-n+3) + \cdots + (n^2+n-3)+(n^2+n-1)\\ & =\frac{n}{2}(n^2-n+1+n^2+n-1)\\ & = n^3. \end{align} This is the sum of $n$ terms of an arithmetic progression.

Answer 3

Note that the first term in each sum $1,3,7,13,21,\cdots$ are quadratic (consider their finite differences and the differences of the differences) thus a formula for this seuence is $n(n-1)+1$. Each sum has $n$ terms and each summand increases by $2$, thus \begin{eqnarray*} \sum_{i=1}^{n}(n(n-1)+2i-1)=n^2(n-1)+n(n+1)-n= \color{red}{n^3}. \end{eqnarray*} So each sum is $n^3$ as observed by lulu in the comments.

Answer 4

Hint: $$ [1+2\cdot( 1+ 2+\ldots +(n-1))]\cdot n + 2(1+2+\ldots +(n-1)) $$

You can then use Gauss formula for $1+ 2+\ldots +(n-1)$.

As in $n=4$: $$13 + (13 + 2) + (13 +4) + (13 +6) = 13\cdot 4 + 2\cdot(1+2+3)$$ and $$ 13 = 1 + 2\cdot( 1+ 2 + 3)$$

Answer 5

We can continue the above pattern as follows: $$ 21 + 23 + 25 + 27 + 29 = 125 = 5^3.$$ $$ 31 + 33 + 35 + 37 + 39 + 41 = 216 = 6^3. $$

We note that the first odd natural number is $1$, which we can write as $$ 1 = 2(1) - 1;$$ the second odd natural number is $3$, which can be written as $$ 3 = 2(2) - 1; $$ the third odd natural number is $5$, which can be written as $$ 5 = 2(3) - 1; $$ and so on and so forth.

We also note that, for each $n \in \mathbb{N}$, at step $n$, we add all the odd natural numbers from the ($ 1 + \sum_{k=1}^n k $)th to the ($\sum_{k=1}^{n+1} k$)th to obtain $n^3$.

Now $$ 1 + \sum_{k=1}^n k = 1 + \frac{n(n+1)}{2} = \frac{n^2 + n + 2}{2}, $$ and $$ \sum_{k=1}^{n+1} k = \frac{ (n+1)(n+2) }{2} = \frac{ n^2 + 3n + 2 }{2}. $$

So we can write $$ \sum_{k= \frac{n^2 + n + 2}{2}}^{\frac{ n^2 + 3n + 2 }{2} } (2k-1) = n^3 $$ for every $n \in \mathbb{N}$.

This can be rewritten as $$ 2 \sum_{k= \frac{n^2 + n + 2}{2}}^{ \frac{ n^2 + 3n + 2 }{2} } k \ - \ \left[ \frac{ n^2 + 3n + 2 }{2} - \frac{n^2 + n + 2}{2} \right] = n^3 $$ for each $n \in \mathbb{N}$.

Or, $$ 2 \sum_{k= \frac{n^2 + n + 2}{2}}^{ \frac{ n^2 + 3n + 2 }{2} } k \ - \ n = n^3 $$ for each $n \in \mathbb{N}$.

Or, $$ 2 \sum_{k= \frac{n^2 + n + 2}{2}}^{ \frac{ n^2 + 3n + 2 }{2} } k = n^3 + n $$ for each $n \in \mathbb{N}$.

We can verify the last identity using mathematical induction.

Hope this works out for you now.

Answer 6

Let $a_n$ the $n$-th term of the sequence.

The sum of the $n$ first terms is the sum of the first $\frac{n(n+1)}{2}$ odd integers, which is $\left( \frac{n(n+1)}{2} \right)^2$.

So that: \begin{equation} a_1+\ldots +a_n = \left( \frac{n(n+1)}{2} \right)^2 \end{equation} \begin{equation} a_1+\ldots +a_{n-1} = \left( \frac{(n-1)n}{2} \right)^2 \end{equation}

Then

$$a_n = \left( \frac{n(n+1)}{2} \right)^2 - \left( \frac{(n-1)n}{2} \right)^2 = \\ \left( \frac{n(n+1)}{2}+\frac{(n-1)n}{2} \right)\left( \frac{n(n+1)}{2}-\frac{(n-1)n}{2} \right) = \left( \frac{n^2+n}{2}+\frac{n^2-n}{2} \right)\left( \frac{n^2+n}{2}-\frac{n^2-n}{2} \right) = n^2 \cdot n = n^3 $$