binomial theoram — MATHS STD 11 Science — Question

Let ${(\sqrt 2 - 1)^6} = f',\,(0 < f' < 1)$,

Since $0 < (\sqrt 2 - 1) < 1$

Now, $k + f + f' = {(\sqrt 2 + 1)^6} + {(\sqrt 2 - 1)^6}$

$ = 2\left\{ {{\,^6}{C_0}{{.2}^3} + {\,^6}{C_2}{{.2}^2} + {\,^6}{C_4}.2 + {\,^6}{C_6}} \right\} = 198$…..(i)

$\therefore \,\,\,f + f' = 198 - k = $ an integer

But, $0 \le f < 1$ and $0 < f' < 1 \Rightarrow 0 < (f + f') < 2$

$ \Rightarrow f + f' = 1$, is an integer

 By (i), $I =198 -(f + f') = 198 -1 = 197$.