Collection of Maths Problems

Sums

Task number: 3312

• Variant

  \(\sum\limits_{i=1}^n i = \frac{1}{2}(n^2+n).\)

• Solution

  For \(n=1\):
  \(\sum\limits_{i=1}^1 i = 1 = \frac{1}{2}(1^2+1).\)

  Induction step \(n-1 \to n\):
  \(\sum\limits_{i=1}^n i = \left(\sum\limits_{i=1}^{n-1} i\right) + n =\frac{1}{2}((n-1)^2+n-1)+n =\frac{1}{2}(n^2-2n+1+n-1+2n) =\frac{1}{2}(n^2+n).\)

• Variant

  \(\sum\limits_{i=1}^n 2i-1 = n^2\).

• Solution

  For \(n=1\):
  \(\sum\limits_{i=1}^1 2i-1 = 1= 1^2\).

  Induction step \(n-1 \to n\):
  \(\sum\limits_{i=1}^n 2i-1= \left(\sum\limits_{i=1}^{n-1} 2i-1\right) + 2n-1 =(n-1)^2+2n-1 =n^2-2n+1+2n-1 =n^2\).

• Variant

  \(\sum\limits_{i=1}^n 4i+5 = 2 n^2+ 7n\).

• Solution

  For \(n=1\):
  \(\sum\limits_{i=1}^1 4i+5 = 9= 2{\cdot} 1^2+ 7{\cdot} 1\).

  Induction step \(n-1 \to n\):
  \(\sum\limits_{i=1}^n 4i+5= \left(\sum\limits_{i=1}^{n-1} 4i+5\right) + 4n+5 =2(n-1)^2+7(n-1)+4n+5 =\\ =2n^2-4n+2+7n-7+4n+5 = 2 n^2+ 7n\)

• Variant

  \(\sum\limits_{i=1}^n i^2 = \frac{1}{3}n^3+\frac{1}{2}n^2+\frac{1}{6}n\).

• Solution

  For \(n=1\):
  \(\sum\limits_{i=1}^1 i^2 = 1= \frac{1}{3}1^3+\frac{1}{2}1^2+\frac{1}{6}\).

  Induction step \(n-1 \to n\):
  \(\sum\limits_{i=1}^n i^2= \left(\sum\limits_{i=1}^{n-1} i^2 \right) + n^2 =\frac{1}{3}(n-1)^3+\frac{1}{2}(n-2)^2+\frac{1}{6}(n-1)+n^2 =\\ =\frac{1}{6}(2n^3-6n^2+6n-2+3n^2-6n+3+n-1+6n^2)=\\ =\frac{1}{6}(2n^3+3n^2+n)= \frac{1}{3}n^3+\frac{1}{2}n^2+\frac{1}{6}n\).

• Variant

  \(\prod\limits_{i=2}^n \frac{i-1}{i} = \frac{1}{n}\).

• Solution

  For \(n=2\):
  \( \prod\limits_{i=2}^2 \frac{i-1}{i}=\frac{1}{2}\).

  Induction step \(n-1 \to n\):
  \(\prod\limits_{i=2}^n \frac{i-1}{i}= \left(\prod\limits_{i=2}^{n-1} \frac{i-1}{i}\right)\frac{n-1}{n} =\frac{1}{n-1}\cdot\frac{n-1}{n}=\frac{1}{n}\).