# What is an = 1*4 + 2*5 + .... +n*(n+3)?

Asked on by gateau66

neela | High School Teacher | (Level 3) Valedictorian

Posted on

We use 1+2^2+3^2+4^2+....+n^2) = n(n+1)(2n+1)/6 and  (1+2+3+4+.....n) = n(n+1)/2.

Consider the general term ar = r*(r+3) = r^2+3r.

Therefore

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

= (1^2+3)+(2^2+2*)+(3^3+3*3)+4^2+4*3)+....(n^2+n*3)

= {1^2+2^2+3^3+4^2)....+n^2}+3(1+2+3+4+...n)

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

= n(n+1){(2n+3)/6 + 3/2}

= n(n+1){2n+1 +9}/6

= n(n+1)(2n+10)/6

= (n+1)(n+1)(n+5)/3.

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

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll write the sum:

Sum k(k + 3), where k is an integer number whose values are from 1 to n.

Sum k(k+3) = Sum (k^2 + 3k)

Sum (k^2 + 3k) = Sum k^2 + Sum 3k

Sum k^2 = 1^2 + 2^2 + ... + n^2 = n(n+1)(2n+1)/6 (1)

Sum 3k = 3*Sum k

Sum k = 1 + 2 + 3 + .... + n = n(n+1)/2

3*Sum k = 3n(n+1)/2 (2)

Sum k(k+3) = (1) + (2)

Sum k(k+3) = n(n+1)(2n+1)/6 + 3n(n+1)/2

We'll factorize by n(n+1)/2:

Sum k(k+3) = [n(n+1)/2]*[(2n+1)/3 + 3]

Sum k(k+3) = [n(n+1)/2]*[(2n + 1 + 9)/3]

Sum k(k+3) = [n(n+1)/2]*[(2n + 10)/3]

Sum k(k+3) = 2*[n(n+1)/2]*[(n + 5)/3]

We'll simplify and we'll get:

Sum k(k+3) = [n(n+1)(n + 5)/3]

So, the value of the general term of the string is:

an = n(n+1)(n + 5)/3

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