1×2+3×4+5×6+7×8······+199×200=

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1×2+3×4+5×6+7×8······+199×200=1×2+3×4+5×6+7×8······+199×200=1×2+3×4+5×6+7×8······+199×200=1×2+3×4+5×

1×2+3×4+5×6+7×8······+199×200=
1×2+3×4+5×6+7×8······+199×200=

1×2+3×4+5×6+7×8······+199×200=
1×2+3×4+5×6+7×8······+199×200=
=1×1 + 1 + 3×3 + 3 + 5×5 + 5 + 7×7 + 7 + …… + 199×199 + 199
= (1² + 3² + 5² + …… + 199²)+ (1+3+5+……+199)
原式也
= 2×2 - 2 + 4×4 - 4 + 6×6 - 6 + …… + 200×200 - 200
= (2² + 4² + 6² + …… + 200²)- (2 + 4 + 6 + …… +200)
因此
原式×2
= (1² + 3² + 5² + …… + 199²)+ (1+3+5+……+199)
+(2² + 4² + 6² + …… + 200²)- (2 + 4 + 6 + …… +200)
= (1²+2²+3²+……+200²)+(1-2+3-4+5-6+……+199-200)
= (1²+2²+3²+……+200²)- 100 【问题就变成我们熟悉的求连续平方和了.可利用公式】
= 200×(200+1)×(200×2+1)/ 6 - 100
= 2686600
因此原式 = 2686600 ÷ 2 = 1343300

oh!no!

#include "stdio.h"
#include "conio.h"
main()
{
long i;
long s=0;
for(i=1;i<=100;i++)
{
s+=(i*2-1)*i*2 ;
}
printf("%ld\n",s);

getch();
}
1343300

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