Question

The arthematic mean of (nC0,nC1,nC2......nCn)?

Answer 1

Arithmetic mean of `C_0^n,C_1^n,C_2^n,....C_n^n` is `2^n/(n+1)`

Explanation:

We know that

`(1+x)^n=C_0^n+C_1^nx+C_2^nx^2+....+C_r^nx^r+....+C_n^nx^n`

Putting `x=1` we get

`C_0^n+C_1^n+C_2^n+....+C_r^n+....+C_n^n=2^n`

Hence arithmetic mean of `C_0^n,C_1^n,C_2^n,....C_n^n` is

`2^n/(n+1)`

Answer 2

`barX=2^n/(n+1)`

Explanation:

`"Using "color(blue)"Binomial theorem:"`

`If ,ninN and x in RR,then`

`color(blue)((1+x)^n=_nC_0+_nC_1* x+_nC_2*x^2+...+_nC_n*x^n`

Take, `x=1`

`(1+1)^n=_nC_0+_nC_1* (1)+_nC_2*(1)^2+...+_nC_n*(1)^n`

`2^n=_nC_0+_nC_1+_nC_2+...+_nC_n`

Note: In expansion of `2^n` there are `color(red)((n+1)` terms.

`"Now "color(violet)"Arithmetic Mean :"`

`color(white)=>color(red)(barX=(sum_(i=1)^nx_i)/N,towhere, N=(n+1)`

`=>barX=(._nC_0+_nC_1+_nC_2+...+_nC_n)/(n+1)`

`=>barX=2^n/(n+1)`