To solve for the sum of the first n number of cubes, use (1+2+3+4+5...+n)2. So for n=100, (1+2+3+4+5...+100)2 = c (cubes) (101 * 50)2 = c 50502 = c c = 25502500
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105625
The sum of the first 100 numbers, excluding zero, is 5,001.
The sum of the first 100 natural numbers is 5,001.
The sum of the first 100 positive even numbers is 10,100.
n*(n+1)=sum 100*(100+1)=10100