=( 1^2+ 1)+(2^2+2)+(3^2+3)+……+(n^2+n)

=( 1^2+2^2+3^2+……+n^2)+( 1+2+3+……+n)

=[n(n+ 1)(2n+ 1)]/6+[n(n+ 1)]

=[n(n+ 1)(5n+2)]/6

1*2+2*3+3*4+ 100* 10 1

= 1^2+ 1+2^2+2+3^2+3+ + 100^2+ 100

= 1^2+2^2+ + 100^2+ 1+2+ + 100

Use the following formula

1^2+2^2+3^2+……+n^2=n(n+ 1)(2n+ 1)/6

1+2+3+ +n=n(n+ 1)/2

In the above formula, n= 100, substituted,

1*2+2*3+3*4+ 100* 10 1= 100* 10 1*20 1/6+ 100* 10 1/2=343400