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Kenneth Rosen Edition 7 Exercise 2.4 Question 37 (Page No. 169)

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Sum both sides of the identity $k^{2}-(k-1)^{2} = 2k-1$ from $k=1$ to $k=n$ and use question $35$ to find

  1. a formula for $\displaystyle{}\sum_{k = 1}^{n}(2k − 1)$ (the sum of the first $n$ odd natural numbers).
  2. a formula for $\displaystyle{}\sum_{k = 1}^{n} k.$
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Kenneth Rosen Edition 7 Exercise 2.4 Question 38 (Page No. 169)
Use the technique given in question $35,$ together with the result of question $37b,$ to derive the formula for $\displaystyle{}\sum_{k=1}^{n}k^{2}$ given in Table $2.\:\:[$Hint: Take $a_{k} = k^{3}$ in the telescoping sum in question $35.]$
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