In a triangular pile of logs, the top row contains one log. Each row below the top row contains one log more than the row above it. If the bottom row contains 435 logs, how many logs are in the pile?

Question

In a triangular pile of logs, the top row contains one log. Each row below the top row contains one log more than the row above it. If the bottom row contains 435 logs, how many logs are in the pile?

Please answer in an organized and neat/easy to read format. And please only include the information that would be understood by a Grade 11. Thank you :)

Answer 1

1st row 1 has 1 log, that's the top row

2nd row 2 has 2 logs

3rd row 3 has 3 logs

...

nth row n has n logs

...

433d row 433 has 433 logs

434th row 434 has 434 logs

435th row 435 has 435 logs, that's the bottom row

there are 435 rows

add up 1+2+3+4+...+ 433+434+435 = 94,830. But that's a little bit tedious and time consuming, even with a calculator or computer

A child prodigy, Carl Gauss, was given a similar problem when he was in about the 3rd grade. The arithmetic instructor wanted to keep him busy for a while and told young Gauss to sum up the numbers 1 through 100. That could take quite a while, especially back then before calculators. He immediately gave the answer = 101(100)/2 = 101(50) = 5050. Do the same with summing 1 through 435. The instructor looked at Gauss very suspiciously. Did he cheat? How did he do it? Gauss had a formula:

sum = (n/2)(n+1) = (435/2)(436) = 218(435) = 94,830 logs in the triangular pile (n= the number of terms = 435)

total logs = the sum of an arithmetic sequence with common denominator d=1 (common denominator = the difference between two successive terms, and first term a1 = 1

Sn = (n/2)(a1 + an) is a formula for the partial sum of an arithmetic sequence.

S435 = (435/2)(1 + 435)

= (435/2)436)

= 94,830 = total logs

(using "logs" as in the example might be a little confusing, or maybe not, as "log" has a very different meaning in math, unrelated to this problem. log is the inverse of an exponent)

Carl Fredrich Gauss outdid every child student for all time. His formula is still used, and no one's come up with anything better.

Answer 2

We have arithmetic sequence: a₁ = 1, d = 1 and n = 435; S₄₃₅ = (1 + 435)·435/2 = 94830