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${(x + 3)^{n - 1}} + {(x + 3)^{n - 2}}(x + 2)$$ + {(x + 3)^{n - 3}}{(x + 2)^2} + ... + {(x + 2)^{n - 1}}$ ના વિસ્તરણમાં ${x^r}[0 \le r \le (n - 1)]$ નો સહગુણક મેળવો.
$^n{C_r}({3^r} - {2^n})$
$^n{C_r}({3^{n - r}} - {2^{n - r}})$
$^n{C_r}({3^r} + {2^{n - r}})$
એકપણ નહિ.
Solution
(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}$
$(\therefore \frac{{{x^n} – {a^n}}}{{x – a}} = {x^{n – 1}} + {x^{n – 2}}{a^1} + {x^{n – 3}}{a^2} + …. + {a^{n – 1}})$
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}})$
