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What is the coefficient of x^5 in the expansion of (2x+3)^7?

Guest Aug 6, 2017
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 #1
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If we think about the expansion of a binomial of a big power (like 7). The expansion looks like the following:
 

\((a+b)^n={n\choose 0}a^nb^0+{n\choose 1}a^{n-1}b^1+...+{n\choose n}a^0b^n\)

 

Knowing that this pattern exists, let's utilize that in the current expansion of the expression:

 

\((2x+3)^7={7\choose0}(2x)^7(3)^0+{7\choose1}(2x)^6(3)^1+{7\choose2}(2x)^5(3)^2+{7\choose3}(2x)^4(3)^3+{7\choose4}(2x)^3(3)^4+...\)

 

Why don't I care about the other terms? Well, I only care about the terms that has the x^5. If it doesn't have that, then I can ignore that term. In this case the appropriate term is \({7\choose 2}(2x)^5(3)^2\). Let's evaluate it:

 

\({7\choose 2}(2x)^5(3)^2\) First, let's evaluate the function that has the 7 choose 2. This has a formula to it.
   

 

The formula for the choose function is the following:
 

\({x \choose y}=\frac{x!}{y!(x-y)!}\)

 

Let's apply that to \({7\choose 2}\) first:

 

\(7\choose2\) Apply the formula to figure out its true value
\(\frac{7!}{2!(7-2)!}\)  
\(\frac{7!}{2!*5!}\) Let's expand the factorial function.
\(\frac{7*6*5*4*3*2*1}{2*1*5*4*3*2*1}\) It is very obvious that there is a common factor of 5! in both the numerator and denominator, so let's factor that out.
\(\frac{7*6}{2}\) Simplify the numerator completely.
\(\frac{42}{2}=21\)  
   

 

Ok, now let's continue with our problem:
 

\({7\choose 2}(2x)^5(3)^2\) Replace \({7\choose 2}\) with its calculated value, 21.
\(21*(2x)^5*3^2\) Distribute the power to the 5 to 2 and the x
\(21*2^5x^5*3^2\) Simplify all the constants.
\(21*32*x^5*9\) Do the multiplication of all the constants.
\(6048x^5\)  
   

 

Therefore, the coefficient in front of the x^5 is 6048.

TheXSquaredFactor  Aug 6, 2017
 #2
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Expand the following:
(2 x + 3)^7

(2 x + 3)^7 = sum_(k=0)^7 binomial(7, k) (2 x)^(7 - k) 3^k = binomial(7, 0) (2 x)^7 3^0 + binomial(7, 1) (2 x)^6 3^1 + binomial(7, 2) (2 x)^5 3^2 + binomial(7, 3) (2 x)^4 3^3 + binomial(7, 4) (2 x)^3 3^4 + binomial(7, 5) (2 x)^2 3^5 + binomial(7, 6) (2 x)^1 3^6 + binomial(7, 7) (2 x)^0 3^7:
128 binomial(7, 0) x^7 + 192 binomial(7, 1) x^6 + 288 binomial(7, 2) x^5 + 432 binomial(7, 3) x^4 + 648 binomial(7, 4) x^3 + 972 binomial(7, 5) x^2 + 1458 binomial(7, 6) x + 2187 binomial(7, 7)

binomial(7, 0) = 1, binomial(7, 1) = 7, binomial(7, 2) = 21, binomial(7, 3) = 35, binomial(7, 4) = 35, binomial(7, 5) = 21, binomial(7, 6) = 7 and binomial(7, 7) = 1:
(2 x)^7 + 7 3 (2 x)^6 + 21 3^2 (2 x)^5 + 35 3^3 (2 x)^4 + 35 3^4 (2 x)^3 + 21 3^5 (2 x)^2 + 7 3^6 2 x + 3^7....................continues............etc.

 

Answer:| 128 x^7 + 1344 x^6 + 6048 x^5 + 15120 x^4 + 22680 x^3 + 20412 x^2 + 10206 x + 2187

Guest Aug 6, 2017

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