How do you solve #(n + 1) ^ { 2} + ( n - 1) ^ { 2} = 2n ^ { 2} + 2#?

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Answer 1 · 2017-01-02T15:32:40

#n# has any value.

Multiply out the brackets first.

Note the identity: #(a+-b)^2 = a^2 +-2ab +b^2#

#(n+1)^2 +(n-1)^2 = 2n^2 +2#

#n^2+2n+1 +n^2 -2n +1 = 2n^2+2#

#2n^2+2= 2n^2 +2" "larr# both sides are identical

#2n^2 -2n^2 = 2-2#

#0=0#

This is a true statement but there is no #n# left to solve for.

This is a special type of equation which is called an identity.

It will be true for any value of the variable.

#n# has any value.

Answer 2 · 2017-01-02T15:39:14

See below.

#(n-1)^2+(n+1)^2=2n^2+2#

is an identity which means that always is true, for any value of #n#

It is like the identity

#cos^2(x)+sin^2(x) = 1#

which is true for any real #x#