Find sum of products of integers 4

We are finding sum of products of consecutive integers.

Last time we had products of 4 integers. Today we have 5.

S = 1(2)3(4)5 + 2(3)4(5)6 + … + n (n+1) (n+2) (n+3) (n+4)

We want to show that S = n (n+1) (n+2) (n+3) (n+4) (n+5)/6 (1)

Check it by Math induction

S(1) = 1(2)3(4)5(6)/6 = 2(3)4(5) = 120

S(n+1) = S(n) + (n+1) (n+2) (n+3) (n+4) (n+5)

= n (n+1) (n+2) (n+3) (n+4) (n+5)/6

• (n+1) (n+2) (n+3) (n+4) (n+5)

= (n+1) (n+2) (n+3) (n+4) (n+5)( n/6 + 1)

= (n+1) (n+2) (n+3) (n+4) ( n + 5) (n+6)/6

By induction equation(1) is correct

Examples: S5 = 5(6)7(8)9(10)/6 = 35(720) = 70(360) = 25200

S10 = 10(11)12(13)14(15)/6 = 110(13)(14)30 = 3300(182) = 600600

Please enjoy and have fun! Thank you!