The coefficient of x^r (0 le r le n - 1) in t | vedclass

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Question

$E = (x + 2)^{n - 1}$ $\left[ {\,1\,\, + \,\,\frac{{x\,\, + \,\,1}}{{x\,\, + \,\,2}}\,\, + \,\,{{\left( {\frac{{x\,\, + \,\,1}}{{x\,\, + \,\,2}}}

Vedclass · Accepted Answer

$^nC_r (2^{n-r} - 1)$

Vedclass · Answer

$E = (x + 2)^{n - 1}$ $\left[ {\,1\,\, + \,\,\frac{{x\,\, + \,\,1}}{{x\,\, + \,\,2}}\,\, + \,\,{{\left( {\frac{{x\,\, + \,\,1}}{{x\,\, + \,\,2}}} ight)}^2}\,\, + \,\,......\,\, + \,\,{{\left( {\frac{{x\,\, + \,\,1}}{{x\,\, + \,\,2}}} ight)}^{n\, - \,1}}\,} ight]$

$= (x + 2)^{n - 1}$ $\left[ {\frac{{1\,\, - \,\,{{\left( {{	extstyle{{x\,\, + \,\,1} \over {x\,\, + \,\,2}}}} ight)}^n}}}{{1\,\, - \,\,{	extstyle{{x\,\, + \,\,1} \over {x\,\, + \,\,2}}}}}} ight]$ 

$= (x + 2)^n$ $\left[ {\frac{{{{(x\,\, + \,\,2)}^n}\,\, - \,\,{{(x\,\, + \,\,1)}^n}}}{{{{(x\,\, + \,\,2)}^n}}}} ight]$ 

$= (x + 2)^n - (x + 1)^n$

Now co-efficient of $x^r$ in $\left[ {\,{{(2\,\, + \,\,x)}^n}\,\, - \,\,{{(1\,\, + \,\,x)}^n}\,} ight] = ^nC_r 2^{n - r} - ^nC_r = ^nC_r (2^{n - r} - 1)$

The coefficient of $x^r (0 \le r \le n - 1)$ in the expression :

$(x + 2)^{n-1} + (x + 2)^{n-2}. (x + 1) + (x + 2)^{n-3} . (x + 1)^2; + ...... + (x + 1)^{n-1}$ is :

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