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Let tr = 2r/2 + 2–r/2 

The Σt2r ∈[r = 1,10] is equal to

(a)  221 - 1 + 20/210

(b)  221 - 1 + 19/210

(c)  221 –1–1/220

(d)  None of these

1 Answer

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Best answer

Correct option   (a)  221 - 1/210  + 19

Explanation:

 ∴  S102

tr2 = 2r + 2–r + 2

= (21+ 2+.........+ 210) + (1/2 + 1/22 + ... + 1/210) + 20

= 211 – 2 + 1 - 1/210 + 20

=  211 - 1/210 + 19

= 221 - 1/210  + 19

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