1Hey Vivek ... Shouldn't the general term be 3√(n*2n*4nn*3n*9n) ???
3√(1.2.4+2.4.8+3.8.12...to n terms/1.3.9+2.6.18+3.12.27+.....to n terms)
plz...i need it urgently that's why i am posting here,otherwise i would post it in 9-10.sry.
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9 Answers
1its general term can b written as 3√n*2n*4n
wich on solving b'cums sigma2n=n(n+1)
1
62\\^3\sqrt{\frac{1.2.4+2.4.8+3.8.12...\text{to n terms}}{1.3.9+2.6.18+3.12.27+.....\text{to n terms}}} \\=^3\sqrt{\frac{1.2.4\left(1.2^0.1+2.2.2+3.2^2.3+...\text{to n terms}\right)}{1.3.9\left(1+2.2.2+3.2^2.3+.....\text{to n terms}\right)}}=^3\sqrt{\frac{1.2.4}{1.3.9}}=2/3
62i think it is one of those questions which should ideally be in class x.. but sometimes we arent able to solve them even in xi-xii ;)