# One equality Common Denominator Form

Problem:

I'm trying to prove the following:

using the following identity:

I have tried various methods, but I am baffled!

Incidentally, I would be very grateful if you could also show me how to prove the second formula.

Solution:

We can solve following identity without using the second one.

First we have to expand the left-hand side term by using the formula

nCr = (n)ï€¡/[(r) ï€¡ (n-r) ï€¡]

Applying in left-hand side we will get

LHS = n-1Ck-m-1*n-1Cm-1 + n-1Ck-m-2*n-1Cm +n-1Ck-m-3*n-1Cm+1 +.......

= (n-1)ï€¡/(k-m-1)ï€¡(n-k+m)ï€¡ * (n-1)ï€¡/(m-1)ï€¡(n-m)ï€¡ + (n-1)ï€¡/(n-k+m+1)ï€¡(k-m-2)ï€¡ *

(n-1)ï€¡/mï€¡(n-1-m)ï€¡ +(n-1)ï€¡/(k-m-3)ï€¡( n-k+m+2)ï€¡ *(n-1)ï€¡/(m+1)ï€¡ (n-m-2)ï€¡ +......

= (n-1)ï€¡*(n-1)ï€¡ { 1/[(k-m-1)ï€¡(n-k+m)ï€¡ (m-1)ï€¡(n-m)ï€¡ ] + 1/[ (n-k+m+1)ï€¡(k-m-2)ï€¡ mï€¡

(n-1-m)ï€¡ ] + 1/[(k-m-3)ï€¡( n-k+m+2)ï€¡ (m+1)ï€¡ (n-m-2)ï€¡] +........

taking (m+n-k)ï€¡ and (k-1)ï€¡ common from denominator

finally we will get

1/[(k-1)ï€¡ (m+n-k)ï€¡]{(n-1)ï€¡ *(n-1+1)(n-1+2)(n-1+3)........(n-1+m)} How does this follow from the previous line?

= 1/[(k-1)ï€¡ (m+n-k)ï€¡]{(m+n-1)ï€¡)}

=(m+n-1)ï€¡/[(k-1)ï€¡ (m+n-k)ï€¡]

= m+n-1Ck-1

https://brainmass.com/math/fractions-and-percentages/one-equality-common-denominator-form-8752

#### Solution Preview

Problem:

I'm trying to prove the following:

using the following identity:

I have tried various methods, but I am baffeled!

Incidentally, I would be very grateful if you could also show me how to prove the second formula.

Solution:

We can solve following identity without using the ...

#### Solution Summary

Various methods are used to solve trigonometric functions. Left-hand side terms by using the formulas are determined.