Explanation
$$Third\>term\>from\>the\>end\>will\>be\>\\(8-3+2)th\>term\>from\>beginning\\T_7=^8C_6(\dfrac{3x}{5})^2(\dfrac{-5}{2x})^6\\=(\dfrac{8\times\>7}{2})\times\>(\dfrac{9}{25})\times x^2\times\>(\dfrac{15625}{64\times\>x^6})\\=(\dfrac{39375}{16x^4})$$
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 = $$27+32 =59$$
Please disable the adBlock and continue. Thank you.