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It is well known that squared brackets do not simply square the individual terms:
(1 + 2)2
6= 12 + 22
(1 + 2 + 3)2
6= 12 + 22 + 32
Instead, we add a correction term ψ to make the equations hold true:
(1 + 2)2 = 12 + 22 + ψ2
(1 + 2 + 3)2 = 12 + 22 + 32 + ψ3
...
(1 + 2 + 3 + ... + n)
2 = 12 + 22 + 32 + ... + n
2 + ψn
Show that the correction term ψn has the following form and determine the values of α and β:
ψn =
n
4 − n
2
α
+
n
3 − n

It is well known that squared brackets do not simply square the individual terms 1 22 6 12 22 1 2 32 6 12 22 32 Instead we add a correction term ψ to make the e class=

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