Collection of Maths Problems

Sums

Task number: 2795

• Variant 1

  \( \displaystyle \sum_{i=1}^n i = \frac{n(n+1)}{2}. \)

• Resolution

  For \(n=1\) the equation \(1=\frac12{\cdot} 1\cdot 2\) holds.

  Assuming that the statement is true for \(n\), we prove it for \(n+1\):

  \( \displaystyle \sum_{i=1}^{n+1}i=\sum_{i=1}^{n}i+n=\frac{n(n+1)}{2}+n=\frac{(n+2)(n+1)}{2} \)

• Variant 2

  \( \displaystyle \sum_{i=1}^n (2i-1) = n^2. \)

• Resolution

  For \(n=1\) the equation \((2-1)=1^2\) holds.

  Assuming that the statement is true for \(n\), we prove it for \(n+1\):

  \( \displaystyle \sum_{i=1}^{n+1}(2i-1)=\sum_{i=1}^{n}(2i-1)+2n+1=n^2+2n+1=(n+1)^2\)

• Variant 3

  \( \displaystyle \sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}6. \)

• Hint

  For \(n=1\) the equation \(1=\frac16{\cdot} 1\cdot 2{\cdot}3\) holds.

  Assuming that the statement is true for \(n-1\), we prove it for \(n\):

  \( \displaystyle \sum_{i=1}^{n}i^2=\sum_{i=1}^{n-1}i^2+n^2=\frac{(n-1)n(2n-1)}6+n^2 =\frac{n(2n^2+3n+1)}6=\frac{n(n+1)(2n+1)}6 \)

• Variant 4

  \( \displaystyle \sum_{i=1}^n i^3 = \left(\frac{n(n+1)}2\right)^2. \)

• Resolution

  For \(n=1\) the equation \(1=\left(\frac12{\cdot} 1\cdot 2\right)^2\) holds.

  Assuming that the statement is true for \(n-1\), we prove it for \(n\):

  \( \displaystyle \sum_{i=1}^{n}i^3=\sum_{i=1}^{n-1}i^3+n^3=\left(\frac{(n-1)n}2\right)^2+n^3 =\frac{n^2(n^2+2n+1)}4=\left(\frac{n(n+1)}2\right)^2 \)