Find The Pattern - 4

It is noted that if the LHS is $n^3$ then the mean or median of the RHS $\mu = n^2$ . Now, we have $\{73, 75 , 77, 79 , \color{#D61F06}{81}, 83, 85, 87, 89\}$ , whose mean or median is $81=9^2$ , therefore, the sum is $9^3 = \boxed{729}$ .

Note:

• For odd $n$ , the mean or median is the center term, for example, $(n, \mu^2): (1,1), (3,9), (5,25), (7,49), (9,81)$ .
• For even $n$ , the mean or median is the mean of the center two terms, for example $(n, \mu^2): (2,4), (4,16), (6,36)$ .

For what is happening:

We note that for $n^3$ , the RHS has $n$ terms, therefore, the mean of RHS is $\dfrac {n^3}n = n^2$ .

Since I am just a 12 year old, ( My bro helped me create this account) I just counted how many numbers were there. 1^3=1 (1 number) 2^3=3+5 (2 numbers) 3^3=7+9+11 (3 numbers) Since 1 number is 1^3, 2 numbers are 2^3, and 3 numbers are 3^3, there is a pattern here. So if there are 9 numbers in the sum, it is 9^3, which is equivalent to 729.

It is an Arithmetic progression (AP) and sum of first n terms S= {n [2a+(n-1)d]/2 } With d=2 , n=9, a=73 S={9 [2×73+(9-1)×2]/2} S=9×(89+73)/2 S=729

Note that ending number is dilating at the rate of 2*n (n: line number). Following that, you'll get the last number of the provided sequence. So, it's easy to predict what's happening at 9 cube .

n^3 = n^2-(n-1) + n^2-(n-1)+2 ...to n. If n=3: 3^3 = 3^2-(3-1) + 3^2-(3-1)+2 + 3^2-(3-1)+4 = 7 + 9 + 11 = 27

Easy: every time, the first number of the addition increases by (2×n)-2 where n is the squared number.

Example: from 1^3=1, 2^3=3+5,

3-1=2.

n is 2, so (2×2)-2=2, correct.

So for every next squared number, the first number of the addition increases by (2×n)-2.

So, 8^3=57,

and 9^3=73 as first numbers.

Problem solved

Not the most honourable way to find a solution but who cares? :D

I just counted on for the series of odd numbers. The first serie of 8 oud numbers equals 8 squared, so the given series is 9 squared.

We note that n^3 = n terms in addition, the lowest being n × (n-1) + 1, which logically always yield an odd number (odd × even = even, +1 = odd) In the question, there is 9 terms, which means it is 9^3. We confirm that 9 × 8 + 1 = 73, the first term.