A cute odd triangle Part 2

Note that the total number of elements in the triangle is the sum of the first 100 integers. Because line 1 has 1 element, line 2 has two, up to 100 on the last line.

Then, using the formula for the sum of the first 100 integers, S=100/2 * (100+1) we get 5050. So the last integer in the last line will be the 5050th odd integer,and the first one in this line will be the 4951th odd integer. Those are, respectively, 10099 and 9901. (you can get them both by using An=1+(n-1)*2, which gives you the n-th odd number)

Now we just have to sum all the odd integers from 9901 to 10099, which we can do by using the same formula used to sum the first 100 integers. S=100/2 * (9901+10099) = 1000000, and we get our result.

@Mayank Raj 's solution is pretty neat and that didn't come to my mind at first , however, I used a different approach which is much more formal and well TBH I like it better:

First off lets see what is the starting number in the 100th row.

Defining $r_n$ as the starting number for row $n$ we have:

$r_1=1 \rightarrow r_2=r_1+2 \rightarrow r_3=r_2+4=r_1+2+4 \rightarrow \dots$

$\rightarrow r_n=r_1+ \displaystyle \sum_{i=1}^{n-1} {2i}$

$Hence \quad \quad r_{100}=1+\displaystyle \sum_{i=1}^{99} {2i} =1+2(1+2+3+\dots+99)=9901$

Now let's see what's the ending number:

Defining $r_e(n)$ as the ending number for the row $n$ we have:

$r_e(1)=r_1=1 \rightarrow r_e(2)=r_2+2 \rightarrow r_e(3)=r_3+2 \times 2$

$\rightarrow \dots \rightarrow r_e(n)=r_n+2(n-1)$

$Hence \quad \quad r_e(100)=9901+2(99)=10099$

It is pretty much obvious the fact that the $n$ th row has $n$ number of elements so now by AM-sequence-summation formula we have:

$S_{100}=\frac {100}{2} \times (9901+10099) =1000000$

A closer look will help in this problem FIRST ROW --- $1$ $=1^{3}$ SECOND ROW --- $3+5= 8$ $=2^{3}$ THIRD ROW--- $7+9+11=27$ $=3^{3}$
SIMILARLY 100TH ROW----SUMMATION OF NUMBERS= $100^{3}$ $\boxed{=1000000}$

100 row = 100 cube

Its obvious that sum of any row equals to n^3.Lets prove it

since sum to n terms is given by

S=n/2 (2a+(n-1)d)

Where,n is no.of terms,a is the first term d is the common difference.

For nth row,

S=n/2 (2a+(n-1)2)

S=n/2 (2a+2n-2)

Now we will find a for nth row

Since for 1st row,a=1=n

2nd row,a=3=n+1

3rd row,a=7=n+(2)^2

4th row=a=13=n+(3)^2

Hence we can create a general formula for a of nth row

a=n+(n-1)^2

Putting this in thr eq of S

Hence,S=n/2 (2 (n+(n-1)^2+2n-2)

S=n/2 (-2n+2n^2 + 2 +2n-2)

S=n^3

which is the general formula for sum of any row

For n=100

S=100^3

s=1000000