Sunday, November 25, 2018

Method of difference

Method of difference

Example 1


$\newcommand{\D}{\displaystyle}$ Find $\D \sum_{k=1}^{n}\frac{1}{k(k+1)}.$
 Let $\D u_k=\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}.$
 \begin{eqnarray} u_1 &=&\frac{1}{1}-\frac{1}{2}\\ u_2 &=&\frac{1}{2}-\frac{1}{3}\\ u_3 &=&\frac{1}{3}-\frac{1}{4}\\ \vdots &=& \vdots\\ u_n &=&\frac{1}{n}-\frac{1}{n+1} \end{eqnarray} By adding, \[u_1+u_2+\cdots+u_n=1-\frac{1}{n+1}\] $\newcommand{\iixi}[2]{\left(\begin{array}{c}#1\\#2\end{array}\right)}$

Proposition: 

For any positive integer $n$ and $k$ \[(n+1)^k-1=\iixi{k}{1}\sum_{i=1}^{n}i^{k-1} +\iixi{k}{2}\sum_{i=1}^{n}i^{k-2}+\cdots+\iixi{k}{k}n .\]

Proof:


 \begin{eqnarray*} (n+1)^k&=&n^k+\iixi{k}{1} n^{k-1}+\iixi{k}{2}n^{k-2}+\cdots+\iixi{k}{k}\\ (n+1)^k-n^k&=&\iixi{k}{1} n^{k-1}+\iixi{k}{2}n^{k-2}+\cdots+\iixi{k}{k}\\ \end{eqnarray*} \begin{eqnarray*} n=1:&2^k-1^k=&\iixi{k}{1} 1^{k-1}+\iixi{k}{2}1^{k-2}+\cdots+\iixi{k}{k}\\ n=2:&3^k-2^k=&\iixi{k}{1} 2^{k-1}+\iixi{k}{2}2^{k-2}+\cdots+\iixi{k}{k}\\ n=3:&4^k-3^k=&\iixi{k}{1} 3^{k-1}+\iixi{k}{2}3^{k-2}+\cdots+\iixi{k}{k}\\ \vdots&\vdots&\vdots\\ &(n+1)^k-n^k=&\iixi{k}{1} n^{k-1}+\iixi{k}{2}n^{k-2}+\cdots+\iixi{k}{k}\\ \end{eqnarray*} By adding \[(n+1)^k-1=\iixi{k}{1}\sum_{i=1}^{n}i^{k-1} +\iixi{k}{2}\sum_{i=1}^{n}i^{k-2}+\cdots+\iixi{k}{k}n \]

Example 2 (Sum of the first $n$ natural numbers) 

 \[S_n=\sum_{i=1}^{n}i=\frac{n(n+1)}{2}.\] By considering $k=2$, in above proposition, we get \[(n+1)^2-1=\iixi{2}{1}\sum_{i=1}^{n}i+\iixi{2}{2}n\] Thus, \begin{eqnarray*} n^2+2n+1-1&=&2S_n+n\\ 2S_n&=&n^2+n=n(n+1)\\ S_n&=&\frac{n(n+1)}{2} \end{eqnarray*}

 Example 3 (Sum of the square of the first $n$ natural numbers, denoted by $S_n^{(2)}$) 


\[S_n^{(2)}=\sum_{i=1}^{n}i^2=\frac{n(n+1)(2n+1)}{6}.\] By considering $k=3$, in above proposition, we get \[(n+1)^3-1=\iixi{3}{1}\sum_{i=1}^{n}i^2+ \iixi{3}{2}\sum_{i=1}^{n}i+\iixi{3}{3}n\] Thus, \begin{eqnarray*} n^3+3n^2+3n+1-1&=&3S_n^{(2)}+3S_n+n\\ n^3+3n^2+3n&=&3S_n^{(2)}+\frac{3n(n+1)}{2}+n\\ 3S_n^{(2)}&=&n^3+3n^2+3n-n-\frac{3n(n+1)}{2}\\ S_n^{(2)}&=&\frac{n(n+1)(2n+1)}{6} \end{eqnarray*}

  

Example 4 (Sum of the square of the first $n$ even numbers) 

\[2^2+4^2+\cdots+(2n)^2=\frac{2n(n+1)(2n+1)}{3}.\] By considering $k=3$, in above proposition, we get \[(n+1)^3-1=\iixi{3}{1}\sum_{i=1}^{n}i^2+ \iixi{3}{2}\sum_{i=1}^{n}i+\iixi{3}{3}n\] Let $u_i=2i,i=1,2,\ldots,n$. Then $(u_i)^2=4i^2.$ Thus, $\D i^2=\frac{(u_i)^2}{4}.$ \begin{eqnarray*} n^3+3n^2+3n+1-1&=&3\sum_{i=1}^{n}\frac{(u_i)^2}{4}+3S_n+n\\ n^3+3n^2+3n&=&\frac{3}{4}\sum_{i=1}^{n}(u_i)^2+\frac{3n(n+1)}{2}+n\\ \frac{3}{4}\sum_{i=1}^{n}(u_i)^2&=&n^3+3n^2+3n-n-\frac{3n(n+1)}{2}\\ \sum_{i=1}^{n}(2i)^2&=&\frac{2n(n+1)(2n+1)}{3} \end{eqnarray*}

Example 5 (Sum of the square of the first $n$ odd numbers) 

\[1^2+3^2+\cdots+(2n-1)^2=\frac{n(2n+1)(2n-1)}{3}\] By considering $k=3$, in above proposition, we get \[(n+1)^3-1=\iixi{3}{1}\sum_{i=1}^{n}i^2+ \iixi{3}{2}\sum_{i=1}^{n}i+\iixi{3}{3}n\] Let $u_i=2i-1,i=1,2,\ldots,n.$ Then $ (u_i)^2=4i^2-4i+1.$ Thus, $\D i^2=\frac{(u_i)^2+4i-1}{4}.$ \begin{eqnarray*} n^3+3n^2+3n+1-1&=&3\sum_{i=1}^{n}\frac{(u_i)^2+4i-1}{4}+3S_n+n\\ n^3+3n^2+3n&=&\frac{3}{4}\sum_{i=1}^{n}(u_i)^2+\frac{3}{4}\sum_{i=1}^{n}(4i-1)+\frac{3n(n+1)}{2}+n\\ \frac{3}{4}\sum_{i=1}^{n}(u_i)^2&=&n^3+3n^2+3n-n-\frac{3n(n+1)}{2}-\frac{3}{4}\sum_{i=1}^{n}(4i-1)\\ \sum_{i=1}^{n}(2i-1)^2&=&\frac{n(2n+1)(2n-1)}{3} \end{eqnarray*}

Example 6 

Show that \[1\times2+2\times3+3\times4\cdots +n(n+1)=\frac{n(n+1)(n+2)}{3}\] Proof: Let $\D u_i=i(i+1)$. By considering $k=3$, in above proposition, we get \[(n+1)^3-1=\iixi{3}{1}\sum_{i=1}^{n}i^2+ \iixi{3}{2}\sum_{i=1}^{n}i+\iixi{3}{3}n\] \begin{eqnarray*} n^3+3n^2+3n+1-1&=&3\sum_{i=1}^{n}(i^2+i)+n\\ n^3+3n^2+3n&=&3\sum_{i=1}^{n}u_i+n\\ 3\sum_{i=1}^{n}u_i&=&n^3+3n^2+3n-n\\ \sum_{i=1}^{n}u_i&=&\frac{1}{3}(n^3+3n^2+2n)\\ &=&\frac{n(n+1)(n+2)}{3} \end{eqnarray*}

Exam Practice 

By using the formula \[1^2+2^2+3^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6},\]
(1)show: \[1^2+3^2+5^2+\cdots+(2n-1)^2=\frac{n(2n+1)(2n-1)}{3}.\]
(2)verify: \[2^2+4^2+6^2+\cdots+(2n)^2=\frac{2n(n+1)(2n+1)}{3}.\]
(3)prove: \[1\times2+2\times3+3\times4\cdots +n(n+1)=\frac{n(n+1)(n+2)}{3}\]
(4) Evaluate: (Product of the same term of an AP) \[(-5)^2+(-2)^2+1^2+4^2+\cdots+ \mbox{10 terms} \] (Hint: Let $\D u_i=-8+3i$. Then $\D u_i^2=\cdots$ )
(5) Find: (Product of the consecutive terms of an AP) \[(-2)(1)+(1)(4)+(4)(7)+\cdots \mbox{10 terms}\] (Hint: Let $\D u_i=(-5+3i)(-2+3i)$)
(6) Calculate: (Product of corresponding  terms of two APs)
\[(-3)(5)+(-1)(8)+(1)(11)+\cdots +\mbox{10 terms}\]


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