1. ## Odd Sequence

A sequence of odd numbers is grouped the following way;

$\displaystyle \{1\},\:\:\{3,5,7\},\:\:\{9,11,13,15,17\},\:\:\{19 ,21,23,25,27,29,31\},\:\:...$

How many numbers are there in the $\displaystyle n$th group?

(this seemed straightforward enough. By observation; $\displaystyle 2n-1$)

How many numbers are there from the $\displaystyle 1$st group to the $\displaystyle (n-1)$th group?

(this is where everything goes downhill for me, would be great to get some help...)

What is the first number in the $\displaystyle n$th group?

What is the sum of all the numbers in the $\displaystyle n$th group?

What is the sum of all the numbers in the $\displaystyle 10$th group?

Once again, any help appreciated.

2. Hello,

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3. Dear davidman,

Number of elements in the $\displaystyle n^{th}$ set= 2n-1

Number of elements from the first set to the $\displaystyle (n-1)^{th}$ set= $\displaystyle \sum_{r=1}^{n-1}(2r-1)=n^2-2n+1$

First element in the $\displaystyle n^{th}$ set= $\displaystyle (n^2-2n+2)^{th} element=1+2(n^2-2n+2-1)=2n^{2}-4n+3$

Sum of all the elements from the first set to the $\displaystyle n^{th}$ set= $\displaystyle \frac{n^2}{2}[2+2(n^2-1)]=n^4$

Sum of all the elements from the first set to the $\displaystyle (n-1)^{th}$ set= $\displaystyle \frac{(n^2-2n+1)}{2}[2+2(n^2-2n+1-1)]=(n^2-2n+1)^2$

Sum of all the elements in the $\displaystyle n^{th}$ set= $\displaystyle n^{4}-(n^2-2n+1)^2$

Sum of all the elements in the $\displaystyle 10^{th}$ set= $\displaystyle 10^4-(10^2-20+1)^2= 3439$

Hope this helps.

4. I translated this problem from my book myself, so I might not be completely accurate with what the original problem is. But here's the answers as given in the back of my book.

$\displaystyle 2n-1$

$\displaystyle (n-1)^2$

$\displaystyle 2n^2-4n+3$

$\displaystyle (2n-1)(2n^2-2n+1)$

$\displaystyle 3439$

5. Dear davidman,

Of course I had miscalculated the last part which I corrected now.