Building Blocks

Note that the nth group contains (2n-1) natural numbers and the last term of the nth group is

1+3+5+....+(2n-1) = $n^{2}$

So, First terms of the nth group is $(n-1)^{2}$ + 1

Thus, sum of first and last term of the nth group is $n^{2}$ + $(n-1)^{2}$ + 1 . In our case , value of n is 100.

Notice that the last term of each group is the square of the group number. $1 = 1 (1^2) \\ 2 = 4 (2^2) \\ 3 = 9 (3^2) \\ \vdots\\ 99 = 9801 (99^2) \\ 100 = 10000 (100^2) \\$ Therefore the first term of the $100th$ group will be $9802$ and the last term $10000.\\ 9802+10000 = \boxed{19802}$ .

First look at the sums of the first and the last term of the first groups. They are as following:

1, 6, 14, 26, 42, ...

As you might notice the difference between the groups from the second group onward follow a certain pattern. So we are going to ignore the first group for now. So we are left with this:

6,14,26,42, ...

As you might notice now the differences between the terms are:

8,12,16,...

So we need the sum of these terms from the first Term till the 98th Term. When we found the result of this, we have the difference between 6 and the Sum of the first and the last Term of the 100th group. So we just need to add 6 and we have the solution. Now we are going to find this sum. To do this we will try to work out a general formula. Therefore we will add the 4 to make it easier. So we have this:

4,8,12,16,...

Now i am going to write the sum till a certain Term beneath the Term. So it looks like this:

4 8 12 16 20 24 ... 4 12 24 40 60 84 ...

now we are going to divide the sums till the nth term by 4. So we get this:

4=4 1 12=4 3 24=4 6 40=4 10 60=6 15 84=4 21

now look at the differences between those results:

1 3 6 10 15 21 2 3 4 5 6

As you see the differences increase by one. Now to we need to find a common formula to find the Sum of the nth Term of the following Sum:

1+2+3+4+5+...

Im going to write the results beneath the sum:

1 2 3 4 5 6
1 3 6 10 15 21

As you see these are also the differences between our previous Set of terms. Now we are going to divide the results by their term so we get the following:

1=1* 1 3=1.5* 2 6=2* 3 10=2.5* 4 15=3* 5
21=3.5* 6

as you can the the factor of the terms always increase by 0.5

After trying around we find the following generla formula to work out the factors:

(n+1)/2

now you just have to multiply this with n to find the sum. So we get:

(n^2+n)/2

As you might recall we added the four to the previous set of Terms although it starts with the 8. We are looking for the 98th term. Therefore we need to do it for the 99th and just subtract the 4, that we added before.

So we get this calculation:

((99^2+99)/2)*4-4

Now we need to add 6, as this is only the difference between 6 and the result:

((99^2+99)/2)*4-4+6

And this gets us our solution.

19802

Please excuse spelling mistakes.

For me, concern the middle member of the nth group since each consequence term has the difference 1

We know the group has the value odd member Suppose the group is $G_n$ and the mid term is $P_n$

$G_1 = (1)$

$G_2 = (2, 3, 4)$

$G_3 = (5, 6, 7, 8, 9)$

$G_4 = (10, 11, 12, 13, 14, 15, 16)$

...

We know that the sum of the first term and the last term is the double of the midterm as long as there are odd members. $a_1 + a_n = 2 a_m, a_m = mid \space term$

Analyse the midterm of each group, we get that:

$P_1 = 1$

$P_2 = 3$

$P_3 = 7$

$P_4 = 13$ ...

We know that there is a progression that the difference of each consequence term is $2k$ and the difference of the difference of each consequence term is exactly constant, 2 so, we know that this progression is \(an^2 + bn +c)

Let's take the first 3 term

\(P_1 = a + b + c = 1 \)

$P_2 = 4a + 2b + c = 3$

$P_3 = 9a + 3b + c = 7$

By substituting and eliminating, we provide $a = 1, b = -1 and c = 1$

So the progression formula is $P_n = n^2 - n +1$

Now, we can look for $P_{100}$

$P_{100} = 100^2 - 100 +1 = 9901 = a_{100}$

Substitute the mid term

$a_1 + a_{199} = 2 a_{100} = 2$

$a_1 + a_{199} = 2 \times 9901= 19802$

It is quite an easy question if you can just find the pattern involved. We can observe that the groups are 1.(1)

2.(2-4)

3.(5-9)

4.(10-16)

5.(17-25)

6.(26-36) And so on..

We observe that the last term of a group is (group no.)^{2}

And the first term of the group is ( group no. - 1)^{2} + 1

So the sum of the first and the last terms of group 100 will be:-

100^2 + ((100-1)^2 + 1)= 19802

Notice that the the relationship between $n$ and $T_n$ where $T_n$ is the first number in group $n$ is $T_n= (n-1)^2+1$

Using this, the first number in group 100 is 9802.

The first number in group 101 is is 10001, so the last number in group 100 is 10000

$10000+9802=19802$

consider an AP where are 100 terms or groups so that 1st term has 1 sub term 2nd term has 2 sub terms and so on where difference is 2 so last terms containing sub terms will have a+ (n-1)d i.e. 1 +99 x 2 =199 then by finding which number will replace last sub term of last term calculate sum of 100 terms i.e. 100/2(2 +99 x 2) =10000 then last sub term of second last term is 99/2(2 + 98 x 2) =9801 so, first sub term of last term is 9802 thus sum will be 9802 + 10000 = 19802