1.If bbb is very small compared to aaa, so that the cube and other higher powers of bbb can be neglected in the identity a1/3b+a2/3(2b)+a(3b)+⋯+an/3(nb)=αn+βn2+γn3a^{1/3}b + a^{2/3}(2b) + a(3b) + \cdots + a^{n/3}(nb) = \alpha n + \beta n^{2} + \gamma n^{3}a1/3b+a2/3(2b)+a(3b)+⋯+an/3(nb)=αn+βn2+γn3, then the value of γ\gammaγ isa.a2+b3a3\dfrac{a^{2} + b}{3a^{3}}3a3a2+bb.b23a3\dfrac{b^{2}}{3a^{3}}3a3b2c.13\dfrac{1}{3}31d.a+b23\dfrac{a + b^{2}}{3}3a+b2Login to continueOnly logged in users canattempt or see the solution.
