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grin007 [14]
3 years ago
10

Please help me solve this​

Mathematics
1 answer:
Viktor [21]3 years ago
3 0

You can factor the 32 out of the sum:

\displaystyle \sum_{m=1}^\infty 32 \cdot \left(\dfrac{1}{2}\right)^{m-1} = 32\sum_{m=1}^\infty \left(\dfrac{1}{2}\right)^{m-1}

We can also change the index as follows

\displaystyle 32\sum_{m=1}^\infty \left(\dfrac{1}{2}\right)^{m-1} = 32\sum_{m=0}^\infty \left(\dfrac{1}{2}\right)^{m}

Now, we have a theorem that states that the series

\displaystyle \sum_{m=1}^\infty a^m

converges if and only if |a|, and in this case we have

\displaystyle \sum_{m=1}^\infty a^m = \dfrac{1}{1-a}

This is your case, because you have

|a|=\dfrac{1}{2}

which implies that your series converges, and the value is

\displaystyle 32\sum_{m=0}^\infty \left(\dfrac{1}{2}\right)^{m} = 32 \cdot \dfrac{1}{1-\frac{1}{2}} = 32\cdot 2 = 64

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Find a number in the closed interval [1/2,3/2] such that the sum of the number and its reciprocal is
sattari [20]

Let's assume that number as x

so,

the sum of the number and its reciprocal is

S(x)=x+\frac{1}{x}

Firstly, we will find derivative

S'(x)=1-\frac{1}{x^2}

now, we can set it to 0

and then we can solve for x

S'(x)=1-\frac{1}{x^2}=0

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Since, only x=1 lies on [1/2,3/2]

so, we will consider only x=1

now, we can plug end values and x=1 into S

At x=1/2:

we can plug x=1/2

S(\frac{1}{2})=\frac{1}{2} +\frac{1}{\frac{1}{2}}

S(\frac{1}{2})=\frac{1}{2} +2

S(\frac{1}{2})=\frac{5}{2}

At x=1:

we can plug x=1

S(1)=1 +\frac{1}{1}

S(1)=1 +1

S(1)=2

At x=3/2:

we can plug x=3/2

S(\frac{3}{2})=\frac{3}{2} +\frac{1}{\frac{3}{2}}

S(\frac{3}{2})=\frac{3}{2} +\frac{2}{3}

S(\frac{3}{2})=\frac{13}{6}

(a)

Smallest value:

The smallest value among them

S(1)=2

So, a number is 1..........Answer

(b)

Largest value:

The largest value among them

S(\frac{1}{2})=\frac{5}{2}

So, a number is \frac{1}{2}........Answer

6 0
4 years ago
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