Given a sequence 1,2,2,3,3,3,4,4,4,4,5,5,5,5,5

[thereddevils]

Homework Statement

Given a sequence 1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,... ie the integer k occurs k times as consecutive terms .

(a) Find the 1989th term

(b) Find the sum of the first 1989 term .

Homework Equations

The Attempt at a Solution

some hints ?

 

[HallsofIvy]

So there are 1 "1", 2 "2"s, 3 "3"s, etc. That means there will be 1+ 2+ 3+ ...+ (n-1) numbers before the first appearence of the number n. That's a well known series, its sum is (1/2)n(n-1). To the 1989th term (that's an old problem!) try to solve (1/2)n(n- 1)= 1989. That's the same as $n(n-1)= n^2- n= 2(1989)= 3978$.

Solve $n^2- n- 3978= 0$ with the quadratic formula. The solution probably will not be an integer, but that will be because you are in the middle of the set of that "n" in the sequence. Take the integer part.

 

[thereddevils]

So there are 1 "1", 2 "2"s, 3 "3"s, etc. That means there will be 1+ 2+ 3+ ...+ (n-1) numbers before the first appearence of the number n. That's a well known series, its sum is (1/2)n(n-1). To the 1989th term (that's an old problem!) try to solve (1/2)n(n- 1)= 1989. That's the same as $n(n-1)= n^2- n= 2(1989)= 3978$.

Solve $n^2- n- 3978= 0$ with the quadratic formula. The solution probably will not be an integer, but that will be because you are in the middle of the set of that "n" in the sequence. Take the integer part.

thanks ! But shouldn't it be 1+2+3+...+n

then , 1/2n(n+1)=1989 and n=63 ?

 

[Unit]

A good hint, I think, is to re-write the series 1 + 2 + 2 + 3 + 3 + 3 + . . . as (1) + (2 + 2) + (3 + 3 + 3) + (4 + 4 + 4 + 4) . . .

Then instead of adding, re-write each bracket as a product: 1(1) + 2(2) + 3(3) + 4(4) + ...

Do you see where to go from here?