Find sum of products of integers 5

We are finding sum of products of consecutive integers.

Last time we had products of 5 integers. Today we have 6.

S = 1(2)3(4)5(6) + 2(3)4(5)6(7) + … + n (n+1) (n+2) (n+3) (n+4) (n+5)

We want to show that S = n (n+1) (n+2) (n+3) (n+4) (n+5) (n+6)/7 (1)

Check it by Math induction

S(1) = 1(2)3(4)5(6) = 2(3)4(5)6 = 720 = 1(2)3(4)5(6)7/7

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

= n (n+1) (n+2) (n+3) (n+4) (n+5) (n+6)/7

• (n+1) (n+2) (n+3) (n+4) (n+5)(n+6)

= (n+1) (n+2) (n+3) (n+4) (n+5)(n+6)( n/7 + 1)

= (n+1) (n+2) (n+3) (n+4) ( n + 5) (n+6)(n+7)/7

By induction equation(1) is correct

Examples: S5 = 5(6)7(8)9(10)11/7 = 30(72)110 = 72(3300) = 237600

S10 = 10(11)12(13)14(15)16/7 = 11(12)(13)300(16) = 3300(2496) = 8236800

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