Answer: (a) \(2^{2}+4^{2}+\ldots+(2n)^{2}=4(1^{2}+2^{2}+\ldots+n^{2})\).

Using \(1^{2}+2^{2}+\ldots+n^{2}=\frac{n(n+1)(2n+1)}{6}\), we get

\(2^{2}+4^{2}+\ldots+(2n)^{2}=4\cdot \frac{n(n+1)(2n+1)}{6}=\frac{2n(n+1)(2n+1)}{3}.\)

(b) \(1^{2}-2^{2}+3^{2}-4^{2}+\ldots-(2n)^{2}\)

\(=(1^{2}+3^{2}+\ldots+(2n-1)^{2})-(2^{2}+4^{2}+\ldots+(2n)^{2}).\)

Now

\(1^{2}+3^{2}+\ldots+(2n-1)^{2}=(1^{2}+2^{2}+\ldots+(2n)^{2})-(2^{2}+4^{2}+\ldots+(2n)^{2}).\)

So

\(1^{2}-2^{2}+3^{2}-4^{2}+\ldots-(2n)^{2}=(1^{2}+2^{2}+\ldots+(2n)^{2})-2(2^{2}+4^{2}+\ldots+(2n)^{2}).\)

Substitute the two sums:

\(=\frac{2n(2n+1)(4n+1)}{6}-2\cdot \frac{2n(n+1)(2n+1)}{3}.\)

Simplifying gives

\(=\frac{n(2n+1)}{3}\big((4n+1)-4(n+1)\big)=-n(2n+1).\)

Hence \(1^{2}-2^{2}+3^{2}-4^{2}+\ldots-(2n)^{2}=-n(2n+1)=-2n^{2}-n.\)

(a) Write the required sum as a multiple of the sum of the first \(n\) squares:

\(2^{2}+4^{2}+\ldots+(2n)^{2}=4(1^{2}+2^{2}+\ldots+n^{2}).\)

Using \(1^{2}+2^{2}+\ldots+n^{2}=\frac{n(n+1)(2n+1)}{6}\),

\(2^{2}+4^{2}+\ldots+(2n)^{2}=4\cdot \frac{n(n+1)(2n+1)}{6}=\frac{2n(n+1)(2n+1)}{3}.\)

(b) Let

\(S=1^{2}-2^{2}+3^{2}-4^{2}+\ldots-(2n)^{2}.\)

Group the odd and even squares:

\(S=(1^{2}+3^{2}+\ldots+(2n-1)^{2})-(2^{2}+4^{2}+\ldots+(2n)^{2}).\)

Also, the odd squares can be found from the sum of all squares minus the even squares:

\(1^{2}+3^{2}+\ldots+(2n-1)^{2}=(1^{2}+2^{2}+\ldots+(2n)^{2})-(2^{2}+4^{2}+\ldots+(2n)^{2}).\)

So

\(S=(1^{2}+2^{2}+\ldots+(2n)^{2})-2(2^{2}+4^{2}+\ldots+(2n)^{2}).\)

Now

\(1^{2}+2^{2}+\ldots+(2n)^{2}=\frac{2n(2n+1)(4n+1)}{6}\)

and from part (a)

\(2^{2}+4^{2}+\ldots+(2n)^{2}=\frac{2n(n+1)(2n+1)}{3}.\)

Hence

\(S=\frac{2n(2n+1)(4n+1)}{6}-2\cdot \frac{2n(n+1)(2n+1)}{3}.\)

Factorising gives

\(S=\frac{n(2n+1)}{3}\big((4n+1)-4(n+1)\big)=\frac{n(2n+1)}{3}(-3)=-n(2n+1).\)

Therefore

\(1^{2}-2^{2}+3^{2}-4^{2}+\ldots-(2n)^{2}=-n(2n+1)=-2n^{2}-n.\)