જો (mathrm{x}+3)^{mathrm{n}-1}+(m | vedclass

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Question

$(x+3)^{n-1}+(x+3)^{n-2}(x+2)+(x+3)^{n-3} $

$ (x+2)^2+\ldots \ldots . .+(x+2)^{n-1} $

$\sum \alpha_r=4^{n-1}+4^{n-2} 	imes 3+4^{n-3} 	imes 3^2

Vedclass · Accepted Answer

$25$

Vedclass · Answer

$(x+3)^{n-1}+(x+3)^{n-2}(x+2)+(x+3)^{n-3} $

$ (x+2)^2+\ldots \ldots . .+(x+2)^{n-1} $

$\sum \alpha_r=4^{n-1}+4^{n-2} 	imes 3+4^{n-3} 	imes 3^2 \ldots \ldots+3^{n-1} $

$=4^{n-1}\left[1+\frac{3}{4}+\left(\frac{3}{4}ight)^2 \ldots . .+\left(\frac{3}{4}ight)^{n-1}ight]$

$=4^{\mathrm{n}-1} 	imes \frac{1-\left(\frac{3}{4}ight)^{\mathrm{n}}}{1-\frac{3}{4}} $

$=4^{\mathrm{n}}-3^{\mathrm{n}}=\beta^{\mathrm{n}}-\gamma^{\mathrm{n}} $

$ \beta=4, \gamma=3 $

$\beta^2+\gamma^2=16+9=25$

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