Let tr = 2^r/2 + 2^–r/2 the Σ tr^2 ; r=1,10 is equal to

Question

Let t_r = 2^r/2 + 2^–r/2

The \(\sum\limits_{r =1}^{10} {t_r}^2\) is equal to

(a) \(\frac{2^{21}-1}{2^{10}} +20\)

(b) \(\frac{2^{21}-1}{2^{10}} +19\)

(c) \(\frac{2^{21}-1}{2^{20}} -19\)

(d) None of these

Answer 1

Correct option is (b) \(\frac{2^{21}-1}{2^{10}} +19\)