Summing, Averaging And Summing Again

Algebra Level 3

f ( n ) = 1 + 2 + 3 + ⋯ + n n \large f(n)=\frac{1+2+3+ \cdots +n}{n}

Consider the summation above for positive integer n n .

Evaluate f ( 1 ) + f ( 2 ) + f ( 3 ) + ⋯ + f ( 99 ) + f ( 100 ) f(1)+f(2)+f(3)+\cdots+f(99)+f(100) .


The answer is 2575.

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Tarmo Taipale by Tarmo Taipale , 22, Finland

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3 solutions

Relevant wiki: Sum of n, n², or n³

f ( n ) = 1 + 2 + 3 + ⋯ + n n = ∑ k = 1 n k n = n ( n + 1 ) 2 n = n + 1 2 \begin{aligned} f(n) & = \frac {1+2+3+\cdots+n}n = \frac {\sum_{k=1}^n k}n = \frac {\frac {n(n+1)}2}n = \frac {n+1}2 \end{aligned}

S = f ( 1 ) + f ( 2 ) + f ( 3 ) + ⋯ + f ( 100 ) = ∑ n = 1 100 n + 1 2 = 1 2 ( ∑ n = 1 100 n + ∑ n = 1 100 1 ) = 1 2 ( 100 ( 101 ) 2 + 100 ) = 2575 \begin{aligned} S & = f(1)+f(2)+f(3)+\cdots + f(100) \\ & = \sum_{n=1}^{100} \frac {n+1}2 \\ & = \frac 12 \left(\sum_{n=1}^{100} n + \sum_{n=1}^{100} 1 \right) \\ & = \frac 12 \left( \frac {100(101)}2 + 100 \right) \\ & = \boxed{2575} \end{aligned}

6 Helpful 0 Interesting 1 Brilliant 0 Confused

Nice! That's what I hoped for. Actually, I think that looks nicer than my version of the solution.

Tarmo Taipale - 4 years, 10 months ago
Lam Nguyen
Aug 6, 2016

f ( 1 ) f(1) = 1 1 \frac{1}{1} = 1 1

f ( 2 ) f(2) = 1 + 2 2 \frac{1+2}{2} = 3 2 \frac{3}{2} = 1.5 1.5

f ( 3 ) f(3) = 1 + 2 + 3 3 \frac{1+2+3}{3} = 6 3 \frac{6}{3} = 2 2

=> 1 1 + 1.5 1.5 + 2 2 + .... + f ( 100 ) f(100)

a a = 1 1

d d = 0.5 0.5

S n S_n = n 2 \frac{n}{2} [ 2 a 2a + d ( n − 1 ) d (n-1) ]

S n S_n = 100 2 \frac{100}{2} [ 2 ∗ 1 2*1 + 0.5 ( 100 − 1 ) 0.5 (100-1) ]

S n S_n = 2575

1 Helpful 0 Interesting 1 Brilliant 0 Confused

Quite good, but it would be even better if you first proved that all f(n) have the same pattern. Which means you have to show that f ( n ) = 1 + ( n − 1 ) 0.5 f(n)=1+(n-1)0.5 . Just assuming it is true, only by looking at the first three members, is not a sure way to solve this problem. It might get you a wrong result, but this time you were lucky.

By the way, you have put 2 below 6 by mistake. (After "f(3)")

Tarmo Taipale - 4 years, 10 months ago

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It seems you have edited it. "6/3" was fixed, but this needs a bit more work. Look at f(n) and simplify it. You can do that by using the arithmetic sum formula. If you don't catch the idea otherwise, look at other solutions.

And when you calculate f(1)+ f(2)+... you use the arithmetic sum formula again, of course, which is already done though.

Tarmo Taipale - 4 years, 10 months ago
Tarmo Taipale
Aug 6, 2016

Relevant wiki: Arithmetic Progression Sum

The formula

∑ k = 1 n ( a 1 + ( k − 1 ) d ) = 2 a 1 + ( k − 1 ) d 2 \sum_{k=1}^n(a_1+(k-1)d)=\frac{2a_1+(k-1)d}{2}

leads in the following result:

f ( n ) = 1 + 2 + 3 + . . . + n n = n ( n + 1 ) 2 n = n + 1 2 = n 2 + 1 2 f(n)=\frac{1+2+3+...+n}{n} = \frac{n(n+1)}{2n} = \frac{n+1}{2} = \frac{n}{2}+\frac{1}{2}

Which means:

f ( 1 ) + f ( 2 ) + f ( 3 ) + . . . + f ( 99 ) + f ( 100 ) = 1 2 + 1 2 + 2 2 + 1 2 + 3 2 + 1 2 + . . . + 100 2 + 1 2 f(1)+f(2)+f(3)+...+f(99)+f(100) = \frac{1}{2}+\frac{1}{2}+\frac{2}{2}+\frac{1}{2}+\frac{3}{2}+\frac{1}{2}+...+\frac{100}{2}+\frac{1}{2}

= 1 + 2 + 3 + . . . + 99 + 100 2 + 100 × 1 2 = 100 × 101 4 + 50 = 2575 = \frac{1+2+3+...+99+100}{2}+100\times\frac{1}{2}=\frac{100\times101}{4}+50=\boxed{2575}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

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