1 + x/2 + x²/4 + ... + xʳ/2ʳ

= (x/2)⁰ + (x/2) + (x/2)² + ... + (x/2)ʳ

This is basically a geometric progression with first term 1 and common ratio x/2.

Sum of the first ten terms means you want up to (x/2)⁹


Using sum of G.P formula Sₙ = a(1 - rⁿ)/(1 - r)


For ten terms it would be :


= 1(1 - (x/2)¹⁰)/(1 - (x/2))

= (1 - x¹⁰/1024)/(1 - x/2)

= (2 - x¹⁰/512)/(2 - x)


(Multiply both numerator and denominator by 2)

For convergence , |r| < 1

So |x/2| < 1

-1 < x/2 < 1

-2 < x < 2

So |x| < 2


Sum to infinity

= a/(1 - r)

= 1/(1 - x/2)

= 2/(2 - x)

(multiply both numerator ans denominator by 2)

Edited