= symbolizes summation$\mathrm{\Sigma}$

first of all , there seems to be a typo in this question

they had it as

, but this is impossible with the other terms listed as a potential series${37}^{{4}}$ it has to be

for it to make sense as a series${37}^{{3}}$

Identify any patterns

each term has a

term being divided by an integer that increases by$37$ each round$1$ there is also a sign flip ( negative to positive ) on each round

They just gave us outright the general form for

.${b}_{k}$ they didn't necessarily have to provide this after the continuation

$\dots $ we could of had to try and determine this structure just by looking at individual terms by themselves in the series.

Find Lower Limit

:$\left(\text{}a\text{}\right)$ the first term in the series they had was

$\left(\text{}1\text{}\right)$ really it was

$\frac{{37}^{0}}{1}$

but they just reduced the first term to just being :

$$\frac{{37}^{1-1}}{1}=\frac{{37}^{0}}{1}=\frac{1}{1}=1$$ the variable renaming gets confusing here

but our first term , or the "starting number" in the series HAS to be 1.

therefore :

Find the Upper Limit

:$\left(\text{}n\text{}\right)$ they sort of gave us this for free

they gave us the equivalent for the last term

$\left(\text{}{b}_{k}\text{}\right)$

$$(-1{)}^{q-1}\ast \frac{{37}^{q-1}}{q}$$ in that last term , the

in our pattern is being raised to$37$ $(\text{}q-1\text{})$ so this last part of the question is really just making sure you understand how sigma notation works in general.

the variable renaming can be confusing here again.

but the summation is written as starting at a value of

,$\left(\text{}a\text{}\right)$ which we know had to be

based on looking at examples in the series$1$ and then it ends at

$\left(\text{}n\text{}\right)$

$$n=q=\mathrm{\infty}$$ just put

for the answer ,$n=q$ but this is how we can relate the concept of a summation series to calculus

if we have some series "at it its limit" or "to infinity"

$$\sum _{x=1}^{\mathrm{\infty}}(\text{}f(x)\text{})$$ more on this later in the "Moving From Series to Calculus" block

Identify any patterns :

each term in the series seems to be in the form of :

so this is the generalized form of

${b}_{k}$

$$\frac{31+k}{k}$$ Find the lower limit

:$\left(\text{}a\text{}\right)$ think how could we get the first term to just be

by itself :$32$ lets assume for a moment

$k=a=1$ that would mean :

Find the upper limit

:$\left(\text{}n\text{}\right)$ the last term in the series was given as :

$$\frac{31+35}{35}$$ you can see it clearly stops at

$35$ therefore ,

$n=35$

Series are used to approximate functions.

some functions are complex , and might be impossible to differentiate.

you can approximate them by using a series.

If the series is finite , like :

Then we can just calculate all of the steps ,

,$x=1$ ,$x=2$ $x=3$ But If the series is going to infinity :

Then we have to take the limit

The derivative of a function = instantaneous rate of change of a function with respect to one of its variables

its also a limit

the derivative of a function at a specific point = the limit of the average rate of change of the function as the interval of the change approaches zero

Its a connection between the processes of differentiation and integration

If you have some function

$f(x)$ And it meets these criteria :

it is continuous on a finite interval , like between

$[\text{}a\text{},\text{}b\text{}]$ its differentiable everywhere on that finite interval

Then , the derivative at any of the potential

values inside the interval is the same as the integral$x$ $${F}^{\prime}(x)=\frac{d}{dx}\ast \left[\text{}{\int}_{a}^{x}(\text{}f(t)\ast dt\text{})\text{}\right]=f(x)$$ is a new function defined by the integral of$F(x)$ The integral

:${\int}_{a}^{x}(\text{}f(t)\ast dt\text{})$ accumulates the area under the curve of the function

from to$a$ $x$

is the variable of integration.$t$ It's a placeholder that represents the current point of evaluation within the interval from

to$a$ $x$

indicates that integration is with respect to$dt$ $t$

Let the anti-derivative , aka the integral of

= $F(x)$ $F(x)={\int}_{a}^{x}f(t)\ast dt$

Then the derivative of

must be just$F(x)$ :$f(x)$ ${F}^{\prime}(x)=f(x)$

via FTOC Part 1 :

we expect to have a variable as one of the limits in the integration

the derivative of an integral with respect to its upper limit ,

is simply the integrand evaluated at that limit

Start with the integral

Apply FTOC Part 1

Let

Then the derivative of

must be just$F(t)$ : $$f(t)=9\ast {e}^{t}$$

Differentiate the integral :

the derivative of

with respect to$F(t)$ is just$t$ so ,

Start with the integral

$${\int}_{1}^{x}(\text{}13\ast ln(\text{}6\ast {u}^{3}\text{})\ast du\text{})$$ Apply FTOC Part 1

Let

Then the derivative of

must be just$F(x)$ :$f(x)$

$$f(x)=13\ast ln(\text{}6\ast {x}^{3}\text{})$$ Differentiate the integral :

the derivative of

with respect to$F(x)$ is just$x$ $f(x)$ so ,

$$\frac{d}{dx}\left[\text{}{\int}_{1}^{x}(\text{}13\ast ln(\text{}6\ast {u}^{3}\text{})\ast du\text{})\text{}\right]=13\ast ln(\text{}6\ast {x}^{3}\text{})$$

Start with the integral

Apply FTOC Part 1

Let

$F(x)={\int}_{0}^{x}(\text{}{(\text{}10\ast {t}^{4}+3\text{})}^{\frac{4}{5}}\ast dt\text{})$ Then the derivative of

must be just$F(x)$ :$f(x)$ $$f(x)={(\text{}10\ast {x}^{4}+3\text{})}^{\frac{4}{5}}$$

Differentiate the integral :

the derivative of

with respect to$F(x)$ is just$x$ $f(x)$ so ,

$$\frac{d}{dx}\left[\text{}{\int}_{0}^{x}(\text{}{(\text{}10\ast {t}^{4}+3\text{})}^{\frac{4}{5}}\ast dt\text{})\text{}\right]={(\text{}10\ast {x}^{4}+3\text{})}^{\frac{4}{5}}$$