Question :- For given identity matrix of order 2, find the value of
[tex]\sf \: 2 I_{2} + 4I_{2} + 8I_{2} + ... + 512I_{2} \\ \\ \\ [/tex]
[tex]\large\underline{\sf{Solution-}}[/tex]
[tex]\sf \: 2 I_{2} + 4I_{2} + 8I_{2} + ... + 512I_{2} \\ \\ [/tex]
We know,
[tex]\sf \: I_{2} = \begin{gathered}\sf\left[\begin{array}{cc}1&0\\ 0&1\end{array}\right]\end{gathered} \\ \\ [/tex]
So, using this value, the above expression can be rewritten as
[tex]\sf \: = 2\begin{gathered}\sf\left[\begin{array}{cc}1&0\\ 0&1\end{array}\right]\end{gathered} + 4\begin{gathered}\sf\left[\begin{array}{cc}1&0\\ 0&1\end{array}\right]\end{gathered} + 8\begin{gathered}\sf\left[\begin{array}{cc}1&0\\ 0&1\end{array}\right]\end{gathered} + ... + 512\begin{gathered}\sf\left[\begin{array}{cc}1&0\\ 0&1\end{array}\right]\end{gathered} \\ \\ [/tex]
[tex]\sf \: = \begin{gathered}\sf\left[\begin{array}{cc}2&0\\ 0&2\end{array}\right]\end{gathered} + \begin{gathered}\sf\left[\begin{array}{cc}4&0\\ 0&4\end{array}\right]\end{gathered} +\begin{gathered}\sf\left[\begin{array}{cc}8&0\\ 0&8\end{array}\right]\end{gathered} + ... + \begin{gathered}\sf\left[\begin{array}{cc}512&0\\ 0&512\end{array}\right]\end{gathered} \\ \\ [/tex]
[tex]\sf = \begin{gathered}\sf\left[\begin{array}{cc}2 + 4 + 8 + ... + 512&0\\ 0&2 + 4 + 8 + ... + 512\end{array}\right]\end{gathered} - - (1)\\ \\ [/tex]
Now, Consider
[tex]\sf \: 2 + 4 + 8 + ... + 512 \\ \\ [/tex]
can be rewritten as
[tex]\sf \: = \: 2 + {2}^{2} + {2}^{3} + ... + {2}^{9} \\ \\ [/tex]
It forms a geometric progression with
First term, a = 2
Common ratio, r = 2
Number of terms, n = 9
So,
[tex]\sf \: = \: \dfrac{2( {2}^{9} - 1)}{2 - 1} \\ \\ [/tex]
[tex]\sf \: = \: 2(512 - 1) \\ \\ [/tex]
[tex]\sf \: = \: 2 \times 511 \\ \\ [/tex]
[tex]\sf \: = \: 1022 \\ \\ [/tex]
So, equation (1) can be rewritten as
[tex]\sf \: = \: \begin{gathered}\sf\left[\begin{array}{cc}1022&0\\ 0&1022\end{array}\right]\end{gathered} \\ \\ [/tex]
[tex]\sf \: = \: 1022\begin{gathered}\sf\left[\begin{array}{cc}1&0\\ 0&1\end{array}\right]\end{gathered} \\ \\ [/tex]
[tex]\sf \: = \: 1022I_{2} \\ \\ [/tex]
Hence,
[tex]\sf \implies \: 2 I_{2} + 4I_{2} + 8I_{2} + ... + 512I_{2} = 1022I_{2}\\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Formulae Used
↝ Sum of n terms of an geometric progression is,
[tex]\begin{gathered}\bigstar\:\:{\underline{{\boxed{\bf{{S_n\:=\dfrac{a( {r}^{n} \: - \: 1)}{r \: - \: 1} \: \rm \: provided \: that \: r \ne \: 1}}}}}} \\ \end{gathered} \\ [/tex]
Wʜᴇʀᴇ,
Sₙ is the sum of n terms of GP.
a is the first term of the sequence.
n is the no. of terms.
r is the common ratio.
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Answers & Comments
Question :- For given identity matrix of order 2, find the value of
[tex]\sf \: 2 I_{2} + 4I_{2} + 8I_{2} + ... + 512I_{2} \\ \\ \\ [/tex]
[tex]\large\underline{\sf{Solution-}}[/tex]
[tex]\sf \: 2 I_{2} + 4I_{2} + 8I_{2} + ... + 512I_{2} \\ \\ [/tex]
We know,
[tex]\sf \: I_{2} = \begin{gathered}\sf\left[\begin{array}{cc}1&0\\ 0&1\end{array}\right]\end{gathered} \\ \\ [/tex]
So, using this value, the above expression can be rewritten as
[tex]\sf \: = 2\begin{gathered}\sf\left[\begin{array}{cc}1&0\\ 0&1\end{array}\right]\end{gathered} + 4\begin{gathered}\sf\left[\begin{array}{cc}1&0\\ 0&1\end{array}\right]\end{gathered} + 8\begin{gathered}\sf\left[\begin{array}{cc}1&0\\ 0&1\end{array}\right]\end{gathered} + ... + 512\begin{gathered}\sf\left[\begin{array}{cc}1&0\\ 0&1\end{array}\right]\end{gathered} \\ \\ [/tex]
[tex]\sf \: = \begin{gathered}\sf\left[\begin{array}{cc}2&0\\ 0&2\end{array}\right]\end{gathered} + \begin{gathered}\sf\left[\begin{array}{cc}4&0\\ 0&4\end{array}\right]\end{gathered} +\begin{gathered}\sf\left[\begin{array}{cc}8&0\\ 0&8\end{array}\right]\end{gathered} + ... + \begin{gathered}\sf\left[\begin{array}{cc}512&0\\ 0&512\end{array}\right]\end{gathered} \\ \\ [/tex]
[tex]\sf = \begin{gathered}\sf\left[\begin{array}{cc}2 + 4 + 8 + ... + 512&0\\ 0&2 + 4 + 8 + ... + 512\end{array}\right]\end{gathered} - - (1)\\ \\ [/tex]
Now, Consider
[tex]\sf \: 2 + 4 + 8 + ... + 512 \\ \\ [/tex]
can be rewritten as
[tex]\sf \: = \: 2 + {2}^{2} + {2}^{3} + ... + {2}^{9} \\ \\ [/tex]
It forms a geometric progression with
First term, a = 2
Common ratio, r = 2
Number of terms, n = 9
So,
[tex]\sf \: = \: \dfrac{2( {2}^{9} - 1)}{2 - 1} \\ \\ [/tex]
[tex]\sf \: = \: 2(512 - 1) \\ \\ [/tex]
[tex]\sf \: = \: 2 \times 511 \\ \\ [/tex]
[tex]\sf \: = \: 1022 \\ \\ [/tex]
So, equation (1) can be rewritten as
[tex]\sf \: = \: \begin{gathered}\sf\left[\begin{array}{cc}1022&0\\ 0&1022\end{array}\right]\end{gathered} \\ \\ [/tex]
[tex]\sf \: = \: 1022\begin{gathered}\sf\left[\begin{array}{cc}1&0\\ 0&1\end{array}\right]\end{gathered} \\ \\ [/tex]
[tex]\sf \: = \: 1022I_{2} \\ \\ [/tex]
Hence,
[tex]\sf \implies \: 2 I_{2} + 4I_{2} + 8I_{2} + ... + 512I_{2} = 1022I_{2}\\ \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Formulae Used
↝ Sum of n terms of an geometric progression is,
[tex]\begin{gathered}\bigstar\:\:{\underline{{\boxed{\bf{{S_n\:=\dfrac{a( {r}^{n} \: - \: 1)}{r \: - \: 1} \: \rm \: provided \: that \: r \ne \: 1}}}}}} \\ \end{gathered} \\ [/tex]
Wʜᴇʀᴇ,
Sₙ is the sum of n terms of GP.
a is the first term of the sequence.
n is the no. of terms.
r is the common ratio.