A) \[^{2m+3}{{C}_{3}}\]
B) \[\frac{1}{3}(m+1)(2{{m}^{2}}+4m+1)\]
C) \[\frac{1}{3}(m+1)(2{{m}^{2}}+4m+3)\]
D) None of these
Correct Answer: C
Solution :
The required number = coefficient of \[{{x}^{2m}}\] in \[{{({{x}^{0}}+{{x}^{1}}+......+{{x}^{m}})}^{4}}\] = coefficient of \[{{x}^{2m}}\] in \[{{\left( \frac{1-{{x}^{m+1}}}{1-x} \right)}^{4}}\] = coefficient of \[{{x}^{2m}}\] in \[{{(1-{{x}^{m+1}})}^{4}}{{(1-x)}^{-4}}\] = coefficient of \[{{x}^{2m}}\] in \[(1-4{{x}^{m+1}}+6{{x}^{2m+2}}+......)\] \[\left( 1+4x+......+\frac{(r+1)(r+2)(r+3)}{3\ !}{{x}^{r}}+.... \right)\] \[=\frac{(2m+1)(2m+2)(2m+3)}{6}-4m\frac{(m+1)(m+2)}{6}\] \[=\frac{(m+1)(2{{m}^{2}}+4m+3)}{3}\].
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