- if - A - n - bmatrix 2 -n 4 -3 -n 3lendbmatrix - - text then - n -1- infty - A - n-A- 0B- 1C- -1D- 1 - 3

The correct option is B 1 |A_n|=3/2^n 4/3^n ∑_n=1^∞|A_n|=3 2=1 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard IX

Mathematics

Multiplication of Matrices

. if ∼ A - n ...

Question

$if A_{n} = [\begin{matrix} 2^{- n} & 4 3^{- n} & 3 \end{matrix}], then \sum_{n = 1}^{\infty} | A_{n} | =$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

-1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1/3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B
1

$| A_{n} | = \frac{3}{2^{n}} - \frac{4}{3^{n}}$
$\sum_{n = 1}^{\infty} | A_{n} | = 3 - 2 = 1$

Suggest Corrections

Similar questions

If A = [\begin{matrix} 2 & - 2 - 2 & 2 \end{matrix}], then A^{n} = 2^{k} A, where k =

I t_{n \to \infty} {\frac{1}{1 + n^{2}} + \frac{2}{2 + n^{2}} + \frac{3}{3 + n^{2}} + \dots_{\dots \dots} + \frac{n}{n + n^{2}}}

Q. If

a_{1}

is the greatest value of

f (x) = \frac{1}{2 + [s i n x]}

and

a_{n + 1} = \frac{(- 1)^{n + 2}}{n + 1} + a_{n}

Then

l i m n \to \infty a_{n} =

_________

If $A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & 1 1 & 1 & 1 1 & 1 & 1 \end{matrix} ⎤ ⎥ ⎦$ , prove that $A^{n} = ⎡ ⎢ ⎣ \begin{matrix} 3^{n - 1} & 3^{n - 1} & 3^{n - 1} 3^{n - 1} & 3^{n - 1} & 3^{n - 1} 3^{n - 1} & 3^{n - 1} & 3^{n - 1} \end{matrix} ⎤ ⎥ ⎦, n \in N$ .

$l i m_{n \to \infty}$ $\frac{n ((2 n + 1)^{2})}{(n + 2) (n^{2} + 3 n - 1)}$ =

Join BYJU'S Learning Program

if An=[2−n43−n3], then ∑∞n=1|An|=

The correct option is B 1 |An|=32n−43n ∑∞n=1|An|=3−2=1

$if A_{n} = [\begin{matrix} 2^{- n} & 4 3^{- n} & 3 \end{matrix}], then \sum_{n = 1}^{\infty} | A_{n} | =$

The correct option is B
1

$| A_{n} | = \frac{3}{2^{n}} - \frac{4}{3^{n}}$
$\sum_{n = 1}^{\infty} | A_{n} | = 3 - 2 = 1$