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alenahelenash · Accepted Answer

To show that the equation1(1−x)2=1+2x+3x2+…+(n+1)xn+n+2−(n+1)x(1−x)2x(n+1)is true, we&#039;ll begin by expanding the right side of the equation and then simplifying it step by step.Let&#039;s start with the expansion:1+2x+3x2+…+(n+1)xn+n+2−(n+1)x(1−x)2x(n+1)To find a common denominator for the terms involving x, we can multiply each term by (1−x)2:(1−x)2+2x(1−x)2+3x2(1−x)2+…+(n+1)xn(1−x)2+(n+2−(n+1)x)x(n+1)Now, let&#039;s simplify each term:(1−x)2=(1−2x+x2)2x(1−x)2=2x(1−2x+x2)=2x−4x2+2x33x2(1−x)2=3x2(1−2x+x2)=3x2−6x3+3x4Continuing this pattern, we can simplify the remaining terms:(n+1)xn(1−x)2=(n+1)xn−2(n+1)xn+1+(n+1)xn+2(n+2−(n+1)x)x(n+1)=(n+2)x(n+1)−(n+1)x(n+2)Now, let&#039;s combine all the simplified terms:(1−2x+x2)+(2x−4x2+2x3)+(3x2−6x3+3x4)+…+((n+1)xn−2(n+1)xn+1+(n+1)xn+2)+((n+2)x(n+1)−(n+1)x(n+2))We can see that the common terms cancel each other out, leaving only the outermost terms:1+(n+2)x(n+1)−(n+1)x(n+2)Now, let&#039;s simplify the terms further:1+(n+2)x(n+1)−(n+1)x(n+2)=1(1−x)2+(n+2)x(n+1)(1−x)2−(n+1)x(n+2)(1−x)2Since all terms have a common denominator, we can combine them into a single fraction:1(1−x)2+(n+2)x(n+1)(1−x)2−(n+1)x(n+2)(1−x)2=1+(n+2)x(n+1)−(n+1)x(n+2)(1−x)2Finally, we can see that the numerator of the fraction is the same as the right side of the original equation. Therefore, we have:1(1−x)2=1+2x+3x2+…+(n+1)xn+n+2−(n+1)x(1−x)2x(n+1)And that concludes the proof.

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