Question

Asking this question third time. now it is a challenge question from me :-

What is the difference in the sum of the squares & the difference of squares of 'n' & 'a' when
\LARGE{\frac{\sqrt{a^{n-2}.3^{a+2}}}{6^n.\left( n \over 2 \right)-\left( 1-{n \over 2} \right)}=\frac{1}{4\sqrt{2^n}}}
& sum of 'a' & 'n' is 12:

Answer 1

\[\LARGE{\frac{\sqrt{a^{n-2}.3^{a+2}}}{6^n.\left( n \over 2 \right)-\left( 1-{n \over 2} \right)}=\frac{1}{4\sqrt{2^n}}}\]

Answer 2

HERE IS THE PROOF

Answer 3

k wait lemme think again :(

Answer 4

it is : \[\large{4\sqrt{2^n}}\] right?

Answer 5

or :
\[\large{4(\sqrt{2})^n}\]

Answer 6

@experimentX , @Callisto @rsadhvika

Answer 7

my work is over now and m totally CONFUSED ,,, :
\[\large{\frac{a^{n-2}*3^{a+2}}{(6^n-1)^2}=2^{-n-2}*\frac{n^2}{4}}\]

Answer 8

@vishweshshrimali5

Answer 9

*

Answer 10

* why?

Answer 11

@jiteshmeghwal9

could u plz tell me what is the source of this problem? (which book? or which competition?)

Answer 12

@mukushla it is the paper of N.T.S.E which is the biggest paper of middle class in india :)