Two ways to find the sum of all adjacent odd even numbers within 1--100

Analysis:

The sum of all adjacent odd even numbers within 100

1 2 3 4 5 6 7 8 9 10 11 ... 100
This is the following:
1*2+3*4+5*6+ ...
Method One:
(2*1-1) 2*1 + (2*2-1) 2*2 + (2*3-1) 2*3 concluded: (2n-1) *2n (50 cycles)
Method Two:

1*2+3*4+...+99*100 concludes: N (n+1) (2 paces per jump i+=2)

#include <stdio.h> #include <stdlib.h>//the sum of all adjacent odd even numbers within 100 and void fun1 () {int res = 0;for (int i=1;i<=50;i++) Loop 50 times {res + = (2*i-1) * (2*i);//Inference Mathematical formula printf ("%d---%d\n", I,res);}} Recommend this, easy to understand void fun2 () {int res = 0;for (int i=1;i<100;i+=2)//Loop 50 times 2 level 2 jumps that output is 1 3 5 7 9 equal odd {res + i* (i+1);//Inference mathematical formula 1*2   3*4   5*6  

Two ways to find the sum of all adjacent odd even numbers within 1--100