1. ## Odd Sequence

A sequence of odd numbers is grouped the following way;

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

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

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

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

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

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

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

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

Once again, any help appreciated.

2. Hello,

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3. Dear davidman,

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

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

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

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

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

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

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

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.

$2n-1$

$(n-1)^2$

$2n^2-4n+3$

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

$3439$

5. Dear davidman,

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