Sum of squares

Algebra Level 5

1 2 + 2 2 + 3 2 + ⋯ + k 2 = ( N − 1 ) N ( N + 1 ) 24 , 1^2+2^2+3^2+\cdots+k^2 = \frac{(N-1)N(N+1)}{24}, where N = 2 k + 1 N = 2k+1 .

Now consider the sum ( 1 2 ) 2 + ( 1 1 2 ) 2 + ( 2 1 2 ) 2 + ⋯ + ( k − 1 2 ) 2 . (\tfrac12)^2+(1\tfrac12)^2+(2\tfrac12)^2+\cdots+(k-\tfrac12)^2. Can this sum also be written in the form ( N − 1 ) N ( N + 1 ) 24 ? \frac{(N-1)N(N+1)}{24}\ ? If so, what is the value of N N ?

No. Yes, N = 2 k + 1 2 N = 2k+\tfrac12 Yes, N = 2 k N = 2k Yes, N = 2 k + 1 N = 2k+1

Arjen Vreugdenhil by Arjen Vreugdenhil , 44, USA

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2 solutions

Method 1: ∑ a = 1 k ( a − 1 2 ) 2 = ∑ a = 1 k ( a 2 − a + 1 4 ) = ∑ a = 1 k a 2 − ∑ a = 1 k a + ∑ a = 1 k 1 4 = ( 2 k ) ( 2 k + 1 ) ( 2 k + 2 ) 24 − k ( k + 1 ) 2 + k 4 = ( 2 k ) ( 2 k + 1 ) ( 2 k + 2 ) − 12 k ( k + 1 ) + 6 k 24 = ( 2 k ) ( 2 k + 2 ) [ ( 2 k + 1 ) − 3 ] + 6 k 24 = ( 2 k ) ( 2 k + 2 ) ( 2 k − 2 ) + 6 k 24 = ( 2 k ) [ ( 4 k 2 − 4 ) + 3 ] 24 = ( 2 k ) ( 4 k 2 − 1 ) 24 = ( 2 k ) ( 2 k − 1 ) ( 2 k + 1 ) 24 = ( N − 1 ) N ( N + 1 ) 24 \sum_{a=1}^k (a-\tfrac12)^2 = \sum_{a=1}^k (a^2 - a + \tfrac14) \\ = \sum_{a=1}^k a^2 - \sum_{a=1}^k a + \sum_{a=1}^k \tfrac14 \\ = \frac{(2k)(2k+1)(2k+2)}{24} - \frac{k(k+1)}2 + \frac k 4 \\ = \frac{(2k)(2k+1)(2k+2) - 12k(k+1) + 6k}{24} \\ = \frac{(2k)(2k+2)\left[(2k+1)-3\right] + 6k}{24} \\ = \frac{(2k)(2k+2)(2k-2) + 6k}{24} \\ = \frac{(2k)\left[(4k^2-4) + 3\right]}{24} \\ = \frac{(2k)(4k^2-1)}{24} \\ = \frac{(2k)(2k-1)(2k+1)}{24} = \frac{(N-1)N(N+1)}{24} with N = 2 k N = 2k .

Method 2: Let the sum be written as S k = ( 1 2 ) 2 + ⋯ + ( k − 1 2 ) 2 S_k = (\tfrac12)^2+\cdots+(k-\tfrac12)^2 . Then 4 S k = 1 2 + 3 2 + ⋯ + ( 2 k − 1 ) 2 = ( 1 2 + 2 2 + 3 2 + 4 2 + ⋯ + ( 2 k − 1 ) 2 + ( 2 k ) 2 ) − ( 2 2 + 4 2 + ⋯ + ( 2 k ) 2 ) = ( 1 2 + ⋯ + ( 2 k ) 2 ) − 4 ⋅ ( 1 2 + ⋯ + k 2 ) = ( 4 k ) ( 4 k + 1 ) ( 4 k + 2 ) 24 − 4 ⋅ ( 2 k ) ( 2 k + 1 ) ( 2 k + 2 ) 24 = k ( 4 k + 1 ) ( 4 k + 2 ) − 2 k ( 2 k + 1 ) ( 2 k + 2 ) 6 = ( 16 k 3 + 12 k 2 + 2 k ) − ( 8 k 3 + 12 k 2 + 4 k ) 6 = 8 k 3 − 2 k 6 = k ( ( 2 k ) 2 − 1 2 ) 3 = k ( 2 k − 1 ) ( 2 k + 1 ) 3 = ( 2 k − 1 ) ( 2 k ) ( 2 k + 1 ) 6 . 4S_k = 1^2 + 3^2 + \cdots + (2k-1)^2 \\ = (1^2 + 2^2 + 3^2 + 4^2 + \cdots + (2k-1)^2 + (2k)^2) - (2^2 + 4^2 + \cdots + (2k)^2) \\ = (1^2 + \cdots + (2k)^2) - 4\cdot (1^2 + \cdots + k^2) \\ = \frac{(4k)(4k+1)(4k+2)}{24} - 4\cdot \frac{(2k)(2k+1)(2k+2)}{24} \\ = \frac{k(4k+1)(4k+2) - 2k(2k+1)(2k+2)}{6} \\ = \frac{(16k^3 + 12k^2 + 2k) - (8k^3 + 12k^2 + 4k)}{6} \\ = \frac{8k^3 - 2k}{6} = \frac{k((2k)^2-1^2)}{3} \\ = \frac{k(2k-1)(2k+1)}{3} = \frac{(2k-1)(2k)(2k+1)}{6}. It follows that S k = ( 2 k − 1 ) ( 2 k ) ( 2 k + 1 ) / 24 S_k = (2k-1)(2k)(2k+1)/24 .

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In method 1:

How we get the second term in the third line :

∑ a = 1 k \ \displaystyle \sum_{a=1}^k a= k ( k + 1 ) 2 \frac {k(k+1)}{2}

Youssef Ali - 5 years, 5 months ago

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In general, ∑ a = 1 k a = 1 + 2 + ⋯ + k = k ( k + 1 ) 2 . \sum_{a=1}^k a = 1+2+\cdots+k = \frac{k(k+1)}2. Check for yourself!

Arjen Vreugdenhil - 5 years, 5 months ago

In the first method it should be ∑ a = 0 k − 1 ( a + 1 2 ) 2 + ( k − 1 2 ) 2 \sum_{a=0}^{k-1} (a+\tfrac12)^2 +(k-\tfrac12)^2

Racchit Jain - 5 years, 5 months ago

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No, because then you wouldn't have the term ( 1 2 ) 2 (\tfrac12)^2 , and you would have the term ( k − 1 2 ) 2 (k-\tfrac12)^2 twice...

Arjen Vreugdenhil - 5 years, 5 months ago

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I have edited my comment so please check now, and how would I have ( k − 1 2 ) 2 (k-\tfrac12)^{2} twice

Racchit Jain - 5 years, 5 months ago

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@Racchit Jain – Because for a = k − 1 a = k-1 , you'd have ( a + 1 2 ) 2 = ( k − 1 + 1 2 ) 2 = ( k − 1 2 ) 2 (a+\tfrac12)^2 = (k-1+\frac12)^2 = (k-\tfrac12)^2 .

The following would be correct ways to write the sum: ∑ a = 0 k − 2 ( a + 1 2 ) 2 + ( k − 1 2 ) 2 ; ∑ a = 0 k − 1 ( a + 1 2 ) 2 ; \sum_{a=0}^{k-2} (a+\tfrac12)^2 + (k-\tfrac12)^2;\ \ \ \ \ \sum_{a=0}^{k-1}(a+\tfrac12)^2; ∑ a = 1 k − 1 ( a − 1 2 ) 2 + ( k − 1 2 ) 2 ; ∑ a = 1 k ( a − 1 2 ) 2 . \sum_{a=1}^{k-1} (a-\tfrac12)^2 + (k-\tfrac12)^2;\ \ \ \ \ \sum_{a=1}^{k}(a-\tfrac12)^2. I find the last one the most useful, but you're welcome to do it differently!

Arjen Vreugdenhil - 5 years, 5 months ago

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@Arjen Vreugdenhil – thanks! i understand now.

Racchit Jain - 5 years, 5 months ago

This is a very nice approach.....however an objective approach: keep k=1 and check with each option.....:P

Samarth Agarwal - 5 years, 4 months ago

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You are right-- that is a quick way to distinguish between the three "Yes" answers. Of course, checking with k = 1 k = 1 does not prove that "Yes" is correct...

Arjen Vreugdenhil - 5 years, 4 months ago
Youssef Ali
Jan 9, 2016

The new sum(L.H.S) is similar to the previous one but the difference is that we subtract 1 2 \frac{1}{2} from every i to k

(If the general term was i 2 i^{2} )

So it can be also written in the same form of N (the sum is different of course)

Now,to find the new value of N :

We put k- 1 2 \frac{1}{2} instead of k in the given equation: N=2k+1

So it will be: N=2(k- 1 2 \frac{1}{2} )+1

N=2k-1+1

N=2k

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No, that reasoning is incorrect... Let's see what happens if we subtract 1 4 \tfrac14 from each term, so that we calculate ( 3 4 ) 2 + ( 1 3 4 ) 2 + ⋯ + ( k − 1 4 ) 2 . (\tfrac34)^2+(1\tfrac34)^2+\cdots+(k-\tfrac14)^2. According to your reasoning, this should be equal to ( N − 1 ) N ( N + 1 ) 24 , \frac{(N-1)N(N+1)}{24}, with N = 2 ( k − 1 4 ) + 1 = 2 k + 1 2 . N = 2(k-\tfrac14)+1 = 2k+\tfrac12. But that is not the case; for instance, in the simple situation k = 1 k = 1 we have N = 2 1 2 N = 2\tfrac12 , and ( N − 1 ) N ( N + 1 ) 24 = ( 1 1 2 ) ( 2 1 2 ) ( 3 1 2 ) 24 = 35 64 ≠ 9 16 = ( 3 4 ) 2 . \frac{(N-1)N(N+1)}{24} = \frac{(1\tfrac12)(2\tfrac12)(3\tfrac12)}{24} = \frac{35}{64} \not= \frac{9}{16} = (\tfrac34)^2.

Arjen Vreugdenhil - 5 years, 5 months ago

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I am sorry but I don't understand why should it be equal to 9 16 \frac {9}{16}

Youssef Ali - 5 years, 5 months ago

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