7 Determine the number of terms: a \( x(1+i) ; x(1+i)^{2} ; x(1+i)^{3} ; \ldots ; x(1+i)^{40} \) b \( x(1+i)^{-1} ; x(1+i)^{-2} ; x(1+i)^{-3} ; \ldots ; x(1+i)^{-160} \) c \( \tan x ; \sin x ; \sin x \cos x ; \ldots ; \tan x(\cos x)^{9} \) d \( 1 ; \sin x ; 1-\cos ^{2} x ; \ldots ;(\sin x)^{11} \)
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For part (a), you're dealing with a sequence that starts at \( x(1+i) \) and goes up to \( x(1+i)^{40} \). Since it begins at the first term and ends at the 40th term, there are a total of 40 terms in this series. In part (b), the progression starts from \( x(1+i)^{-1} \) and continues to \( x(1+i)^{-160} \). This creates terms that represent a sequence from \( -1 \) to \( -160 \). To find the number of terms in this case, you would calculate \( 160 - 1 + 1 \), which gives you 160 terms. For part (c), the sequence begins with \( \tan x \) and progresses through \( \sin x \) and \( \sin x \cos x \) and ends with \( \tan x (\cos x)^{9} \). If we observe the structure, the terms seem to be related to alternating trigonometric functions and their combinations but can be seen to have a power of 9 in their final term, suggesting an organized sequence that totals 10 terms. Finally, part (d) starts from 1 and goes to \( (\sin x)^{11} \). Since you're moving through terms by increasing the power of sine function from 0 to 11, the total count from \( 1 \) (considered as \( (\sin x)^{0} \)) to \( (\sin x)^{11} \) gives you 12 terms in this case.
