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Question: \[\frac{1}{4}\sum_{k=3}^{n}\frac{(k-1)^{2}(k-2)^{2}}{k^{2}}\]...

14∑k=3n(k−1)2(k−2)2k2\frac{1}{4}\sum_{k=3}^{n}\frac{(k-1)^{2}(k-2)^{2}}{k^{2}}

Answer
14[n(n+1)(2n+1)6−5  −  6(n(n+1)2−3)  +  13(n−2)  −  12(Hn−32)  +  4∑k=3n1k2]\frac{1}{4}\Bigg[\frac{n(n+1)(2n+1)}{6}-5 \;-\;6\Big(\frac{n(n+1)}{2}-3\Big)\;+\;13(n-2)\;-\;12\bigl(H_{n}-\tfrac{3}{2}\bigr)\;+\;4\sum_{k=3}^{n}\frac{1}{k^{2}}\Bigg]
Explanation

Solution

Step 1. Simplify the general term:

(k−1)2(k−2)2k2=((k−1)(k−2)k)2=(k−3+2k)2=k2−6k+13−12k+4k2.\frac{(k-1)^{2}(k-2)^{2}}{k^{2}} =\biggl(\frac{(k-1)(k-2)}{k}\biggr)^{2} =\bigl(k-3+\tfrac{2}{k}\bigr)^{2} =k^{2}-6k+13-\frac{12}{k}+\frac{4}{k^{2}}.

Step 2. Split the sum:

∑k=3n(k2−6k+13−12k+4k2)=∑k2−6∑k+13∑1−12∑1k+4∑1k2.\sum_{k=3}^{n}\bigl(k^{2}-6k+13-\tfrac{12}{k}+\tfrac{4}{k^{2}}\bigr) =\sum k^{2}-6\sum k+13\sum1-12\sum\tfrac1k+4\sum\tfrac1{k^{2}}.

Step 3. Use standard results:

  • ∑k=3nk2=n(n+1)(2n+1)6−5\sum_{k=3}^{n}k^{2}=\frac{n(n+1)(2n+1)}{6}-5
  • ∑k=3nk=n(n+1)2−3\sum_{k=3}^{n}k=\frac{n(n+1)}{2}-3
  • ∑k=3n1=n−2\sum_{k=3}^{n}1=n-2
  • ∑k=3n1k=Hn−1−12=Hn−32\sum_{k=3}^{n}\frac1k=H_{n}-1-\tfrac12=H_{n}-\tfrac{3}{2}
  • ∑k=3n1k2\sum_{k=3}^{n}\frac1{k^{2}} remains as is or can be written =ζ(2)−1−14−∑k>n1/k2.=\zeta(2)-1-\tfrac14-\sum_{k>n}1/k^{2}.

Step 4. Multiply the total by 14\tfrac14 to obtain the final closed‑form:

14[n(n+1)(2n+1)6−5  −  6(n(n+1)2−3)  +  13(n−2)  −  12(Hn−32)  +  4∑k=3n1k2].\boxed{ \frac{1}{4}\Bigg[\frac{n(n+1)(2n+1)}{6}-5 \;-\;6\Big(\frac{n(n+1)}{2}-3\Big)\;+\;13(n-2)\;-\;12\bigl(H_{n}-\tfrac{3}{2}\bigr)\;+\;4\sum_{k=3}^{n}\frac{1}{k^{2}}\Bigg]. }