Question

Fun Little Binomial question<3

Answer 1

If \[\sum_{r=0}^{3n}a_r(x-4)^r=\sum_{r=0}^{3r}A_r(x-5)^r\] and \(a_k=1~\forall~K \ge 2n\) and \[\sum_{r=0}^{3n}d_r(x-8)^r=\sum_{r=0}^{3n}B_r(x-9)^r\] and \[\sum_{r=0}^{3r}d_r(x-12)^r=\sum_{r=0}^{3r}D_r(x-13)^r \] and \(d_k=1 ~\forall ~K \ge 2n\). Then find the value of \(\Large \frac{A_{2n}+D_{2n}}{B_{2n}}\).

Answer 2

Are you sure there are no typoes ?

Answer 3

@imqwerty

Answer 4

i checked :) there are no typos this time (B

Answer 5

We can do appropriate substitutions because the question in its current form doesn't need to be so complicated.\[\sum_{r=0}^{3n}a_r(1+t)^r=\sum_{r=0}^{3n}A_rt^r\]The rest of the questions will also have the same expression (we're also given that \(d_{2n}=a_{2n}\), so for a start, we know that \(a_r=d_r\), and \(A_{2n} = B_{2n} = D_{2n}\). Therefore \(\frac{A_{2n}+B_{2n}}{D_{2n}}=2\).

Answer 6

I hurried through the question this time. My answer may not be right.

Answer 7

parth is correct!!~ :)

Answer 8

Oh God that answer is badly written. \(a_r = d_r\) for all \(r\) is not necessary as I wrote above.

Answer 9

ugh