Prove that: #(s-a_1)^2 + (s-a_2)^2 +cdots+ (s-a_n)^2 =a_1^2 +a_2^2+cdots+a_n^2# when #a_1^2 +a_2^2+cdots+a_n^2 = n/{2s}#?

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Answer 1 · 2017-09-27T23:05:08

The given identity is false, but if:

#a_1+a_2+...+a_n = n/2 s#

then:

#(s-a_1)^2+(s-a_2)^2+...+(s-a_n)^2 = a_1^2+a_2^2+...+a_n^2#

Consider the case #n=1#, #s=2#, #a_1 = 1/2#

Then:

#a_1^2 = (1/2)^2 = 1/4 = 1/(2*2) = n/(2s)#

#(s-a_1)^2 = (2-1/2)^2 = (3/2)^2 = 9/4 != 1/4 = a_1^2#

So something is not quite right with the question as stands.

Bonus

Let's try to wo##rk out the correct condition.

Working back from the desired result:

#(s-a_1)^2+(s-a_2)^2+...+(s-a_n)^2 = a_1^2+a_2^2+...+a_n^2#

Multiplying out the left hand side, we have:

#a_1^2+a_2^2+...+a_n^2-2s(a_1+a_2+...+a_n)+ns^2 = a_1^2+a_2^2+...+a_n^2#

Subtract #a_1^2+a_2^2+...+a_n^2# from both sides to get:

#-2s(a_1+a_2+...+a_n)+ns^2 = 0#

Add #2s(a_1+a_2+...+a_n)# to both sides to get:

#ns^2 = 2s(a_1+a_2+...+a_n)#

Divide both sides by #2s# to get:

#n/2 s = a_1+a_2+...+a_n#

Note that all of the above steps are reversible, so if:

#a_1+a_2+...+a_n = n/2 s#

then:

#(s-a_1)^2+(s-a_2)^2+...+(s-a_n)^2 = a_1^2+a_2^2+...+a_n^2#