Question 977126
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We can eliminate a for the reason Alan gave.


Then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 7^{140}\ =\ (8\ -\ 1)^{140}\ =\ (2^3\ -\ 1)^{140}\ =\ 2^{420}\ -\ 140\cdot 2^{417}\ +\ ...]


By the binomial theorem, and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 17^{105}\ =\ (16\ +\ 1)^{105}\ =\ (2^4\ -\ 1)^{105}\ =\ 2^{420}\ +\ 105\cdot 2^{416}\ +\ ...]


Notice that in the expansion of *[tex \LARGE  (8\ -\ 1)^{140}], the signs alternate, but in the expansion of *[tex \LARGE (16\ +\ 1)^{105}] the signs are all positive.  Therefore


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 17^{105}\ >\ 7^{140}]


Similarly, the first term of *[tex \Large 31^{84}\ =\ (32\ -\ 1)^{84}] will be *[tex \Large 2^{420}] and the terms will alternate, hence


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 17^{105}\ >\ 31^{84}]


Therefore, answer c, as Alan's calculator confirms.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \