Sequences and Series — MATHS STD 11 Science — Question

1. 64

Solution:

$32\times32^{\frac{1}{6}}\times32^{\frac{1}{36}}\times\ \cdots\infty$

$=32^{\big(1+\frac{1}{6}+\frac{1}{36}+\ \cdots\infty\big)}$

$=32^{\Bigg(\frac{1}{1-\frac{1}{6}}\Bigg)}$ $[\because$ it is a G.P. $]$

$=32^{\big(\frac65\big)}$

$=\big(2^5\big)^{\big(\frac65\big)}$

$=2^6$

$=64$