Binomial theorem

Question

IF α=(5/2!*3)+ 5*7/3!*3²+5*7*9/4!*3³.....THE FIND VALUE OF α²+4α

Answer 1

If this is written like this (  a= [5/(2! * 3)] + [(5*7)/(3! * 32)] + [(5*7*9) / (4! * 33)]    ),  then a will be (1695/648) or (565/216).  Then a²+4a will be:

 

(565/216)²  + (565/162)  = (3 * 565) + (54 * 4 * 4 * 565)

                                                         (216^₂ * 3)

So, LCD is (3 * 216²) or (3 * 54 * 4 * 54 * 4) = 139968

So, if you add two numerators  (3 * 565) + (864 * 565) = 565 (3 + 864) = 565 * 867 = 489855 or leave it as (565*867).

 

So, rewrite it.  565*867

                     (216)² * 3

 

Let's break it down the numerator.  565 = 5*113.       867 = 3 * 289 = 3 * 17 * 17

So, we have 5 * 113 * 3 * 17 * 17.

 

Let's break it down the denominator.  216 * 216 * 3 = 36 * 6 * 36 * 6 * 3

So, we have 36 * 6 * 36 * 6 * 3.

 

So, rewrite again.

 

5 * 3 * 17 * 17 * 113

36 * 6 * 36 * 6 * 3

 

Only thing we can cancel is 3.  So rewrite again.

 

5 * 17 * 17 * 113

36 * 6 * 36 * 6

 

which is (163285 / (216)²).   or (163285 / 46656).

 

Yes, it is long and tedious.  I hope you understand what I just showed you.

Good luck.