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) \(\frac{2^{21}-1+20}{2^{10}}\)

(b) \(\frac{2^{21}-1+19}{2^{10}}\)

(c) \(\frac{2^{21}-1-1}{2^{20}}\)

(d) None of these

Answer 1

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

tr² = 2^r + 2^–r + 2