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nCr + nCr-1 = n!/[r!(n-r)!] + n!/[(r-1)!(n-r+1)!]

= n!/[(r-1)!(n-r)!]*{1/r + 1/n-r+1}

= n!/[(r-1)!(n-r)!]*{[(n-r+1) + r]/[r*(n-r+1)]}

= n!/[(r-1)!(n-r)!]*{(n+1)/r*(n-r+1)]}

= (n+1)!/[r!(n+1-r)!]

= n+1Cr

This answer is:
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nCr + nCr-1 = n!/[r!(n-r)!] + n!/[(r-1)!(n-r+1)!]

= n!/[(r-1)!(n-r)!]*{1/r + 1/n-r+1}

= n!/[(r-1)!(n-r)!]*{[(n-r+1) + r]/[r*(n-r+1)]}

= n!/[(r-1)!(n-r)!]*{(n+1)/r*(n-r+1)]}

= (n+1)!/[r!(n+1-r)!]

= n+1Cr

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You would take the following steps for G = A / (1-R):

G = A / (1-R)

Multiply by (1-R):

G * (1-R) = A

Divide by G:

(1-R) = A/G

1-R = A/G

Subtract 1:

-R = (A/G) - 1

Divide By -1:

R = -((A/G) - 1)

Check Work:

Original Problem:

A = 12; R = 5

G = 12 / (1-5)

G = -3

Solving For R:

R = -((12/-3)-1)

R = 5

Therefore, R= -((A/G)-1)

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Consider a denominator of r;

It has proper fractions:

1/r, 2/r, ...., (r-1)/r

Their sum is: (1 + 2 + ... + (r-1))/r

The numerator of this sum is

1 + 2 + ... + (r-1)

Which is an Arithmetic Progression (AP) with r-1 terms, and sum:

sum = number_of_term(first + last)/2

= (r-1)(1 + r-1)/2

= (r-1)r/2

So the sum of the proper fractions with a denominator or r is:

sum{r} = ((r-1)r/2)/r = ((r-1)r/2r = (r-1)/2

Now consider the sum of the proper fractions with a denominator r+1:

sum{r+1} = (((r+1)-1)/2

= ((r-1)+1)/2

= (r-1)/2 + 1/2

= sum{r) + 1/2

So the sums of the proper fractions of the denominators forms an AP with a common difference of 1/2

The first denominator possible is r = 2 with sum (2-1)/2 = ½;

The last denominator required is r = 100 with sum (100-1)/2 = 99/2 = 49½;

And there are 100 - 2 + 1 = 99 terms to sum

So the required sum is:

sum = ½ + 1 + 1½ + ... + 49½

= 99(½ + 49½)/2

= 99 × 50/2

= 2475

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Resistors in parallel equation is

!/R = 1/r(1)_ + 1/r)2) + 1/r(3)

Since they are all 8 ohms.

Then

1/R = 1/8 + 1/8 + 1/8

1/R = 3/8

R = 8/3

R = 2 2/3 = 2.6666....

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