Coefficients of {x^r}[0 le r le (n - 1)] in t | vedclass

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Question

(b) We have ${(x + 3)^{n - 1}} + {(x + 3)^{n - 2}}(x + 2) + $ ${(x + 3)^{n - 3}}{(x + 2)^2} + .... + {(x + 2)^{n - 1}}$

$ = \frac{{{{(x + 3)}^n} -

Vedclass · Accepted Answer

$^n{C_r}({3^{n - r}} - {2^{n - r}})$

Vedclass · Answer

(b) We have ${(x + 3)^{n - 1}} + {(x + 3)^{n - 2}}(x + 2) + $ ${(x + 3)^{n - 3}}{(x + 2)^2} + .... + {(x + 2)^{n - 1}}$

$ = \frac{{{{(x + 3)}^n} - {{(x + 2)}^n}}}{{(x + 3) - (x + 2)}} = {(x + 3)^n} - {(x + 2)^n}$ 

$(	herefore \frac{{{x^n} - {a^n}}}{{x - a}} = {x^{n - 1}} + {x^{n - 2}}{a^1} + {x^{n - 3}}{a^2} + .... + {a^{n - 1}})$

Therefore coefficient of ${x^r}$ in the given expression

= Coefficient of ${x^r}$ in $[{(x + 3)^n} - {(x + 2)^n}]$

$ = {\,^n}{C_r}{3^{n - r}} - {\,^n}{C_r}{2^{n - r}} = {\,^n}{C_r}({3^{n - r}} - {2^{n - r}})$

Coefficients of ${x^r}[0 \le r \le (n - 1)]$ in the expansion of ${(x + 3)^{n - 1}} + {(x + 3)^{n - 2}}(x + 2)$$ + {(x + 3)^{n - 3}}{(x + 2)^2} + ... + {(x + 2)^{n - 1}}$

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