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  1. Problems
  2. /Find the sum of all the coefficients, and the sum with signs alternating, in the expansion of (3x +…
EasyMathematicsHigh School (11–12)Algebra › Binomial TheoremNumericalA-LevelIB

Find the sum of all the coefficients, and the sum with signs alternating, in the expansion of (3x +…

StemPy LabContributed by StemPy Lab

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

Find the sum of all the coefficients, and the sum with signs alternating, in the expansion of (3x+2)4(3x + 2)^4(3x+2)4.

Check your answer

Work it out on paper, then type your number. The answer itself stays hidden.

Need a reminder? The key ideaShowHide

The coefficients of (1+x)n(1 + x)^n(1+x)n add up to 2n2^n2n (put x=1x = 1x=1), and the odd-placed and even-placed terms each add up to 2n−12^{n - 1}2n−1 (use x=−1x = -1x=−1). For a polynomial fff, the even powers add up to f(1)+f(−1)2\dfrac{f(1) + f(-1)}{2}2f(1)+f(−1)​ and the odd powers to f(1)−f(−1)2\dfrac{f(1) - f(-1)}{2}2f(1)−f(−1)​. Also ∑r⋅nCr=n⋅2n−1\sum r \cdot {}^{n}C_{r} = n \cdot 2^{n - 1}∑r⋅nCr​=n⋅2n−1 and ∑nCrr+1=2n+1−1n+1\sum \dfrac{{}^{n}C_{r}}{r + 1} = \dfrac{2^{n + 1} - 1}{n + 1}∑r+1nCr​​=n+12n+1−1​.

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Step 1. The total. Put x=1x = 1x=1: (3+2)4=54=625(3 + 2)^4 = 5^4 = 625(3+2)4=54=625.

Step 2. The alternating sum. Put x=−1x = -1x=−1: (−3+2)4=(−1)4=1(-3 + 2)^4 = (-1)^4 = 1(−3+2)4=(−1)4=1.

Answer: 625625625 and 111.

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