Explanation
Thirdtermfromtheendwillbe(8−3+2)thtermfrombeginningT7=8C6(3x5)2(−52x)6=(8×72)×(925)×x2×(1562564×x6)=(3937516x4)
Given that ,
{{\left( {{3}^{\frac{1}{5}}}+{{2}^{\frac{1}{3}}} \right)}^{15}}
General term is
{{t}_{r+1}}{{=}^{15}}{{C}_{r}}{{\left( {{3}^{\frac{1}{5}}} \right)}^{15-r}}{{\left( {{2}^{\frac{1}{3}}} \right)}^{r}}
{{=}^{15}}{{C}_{r}}{{.3}^{\frac{15-r}{5}}}{{.2}^{\frac{r}{3}}}
For general term r=0,5
{{t}_{0+1}}{{=}^{15}}{{C}_{0}}{{.3}^{3}}=27
{{t}_{15+1}}={{t}_{16}}{{=}^{15}}{{C}_{15}}{{3}^{0}}.{{2}^{5}} =32
Hence, this is the required value = 27+32 =59
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