Expand (3x-4y)^11: Using Pascal's Triangle

by Alex Johnson 43 views

Expanding binomials, especially those raised to higher powers, can seem daunting. However, Pascal's Triangle provides a neat and efficient method to achieve this. In this article, we'll explore how to use Pascal's Triangle to expand the binomial (3x−4y)11(3x - 4y)^{11}.

Understanding Pascal's Triangle

Pascal's Triangle is a triangular array of numbers where each number is the sum of the two numbers above it. The triangle starts with a 1 at the top, and each subsequent row is constructed based on the previous one. The numbers in Pascal's Triangle are the binomial coefficients, which are essential for expanding binomial expressions.

Constructing Pascal's Triangle

To construct Pascal's Triangle, begin with a 1 at the top. Each row starts and ends with 1. The other numbers in the row are the sum of the two numbers directly above them. Here are the first few rows:

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
...

Each row corresponds to the coefficients in the binomial expansion of (a+b)n(a + b)^n, where nn is the row number (starting with n=0n = 0 for the top row).

Binomial Theorem and Pascal's Triangle

The Binomial Theorem states that for any non-negative integer nn:

(a+b)n=∑k=0n(nk)an−kbk(a + b)^n = \sum_{k=0}^{n} {n \choose k} a^{n-k} b^k

Where (nk){n \choose k} represents the binomial coefficient, which can be found in Pascal's Triangle. Specifically, (nk){n \choose k} is the (k+1)(k+1)-th entry in the (n+1)(n+1)-th row of Pascal's Triangle (remembering that we start counting rows and entries from 0).

Expanding (3x−4y)11(3x - 4y)^{11} using Pascal's Triangle

Now, let's apply Pascal's Triangle to expand (3x−4y)11(3x - 4y)^{11}. First, we need to find the coefficients from the 12th row of Pascal's Triangle (since the first row is row 0).

The 12th row of Pascal's Triangle is:

1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1

These numbers will be the coefficients in our expansion. Now, we apply the binomial theorem:

(3x−4y)11=∑k=011(11k)(3x)11−k(−4y)k(3x - 4y)^{11} = \sum_{k=0}^{11} {11 \choose k} (3x)^{11-k} (-4y)^k

Let's write out the terms:

  1. ${11 \choose 0} (3x)^{11} (-4y)^0 = 1 \cdot (3x)^{11} \cdot 1 = 177147x^{11}
  2. (111)(3x)10(−4y)1=11⋅(3x)10⋅(−4y)=11⋅59049x10⋅(−4y)=−2598156x10y{11 \choose 1} (3x)^{10} (-4y)^1 = 11 \cdot (3x)^{10} \cdot (-4y) = 11 \cdot 59049x^{10} \cdot (-4y) = -2598156x^{10}y
  3. (112)(3x)9(−4y)2=55⋅(3x)9⋅(16y2)=55⋅19683x9⋅(16y2)=17372160x9y2{11 \choose 2} (3x)^{9} (-4y)^2 = 55 \cdot (3x)^{9} \cdot (16y^2) = 55 \cdot 19683x^{9} \cdot (16y^2) = 17372160x^{9}y^2
  4. (113)(3x)8(−4y)3=165⋅(3x)8⋅(−64y3)=165⋅6561x8⋅(−64y3)=−69672960x8y3{11 \choose 3} (3x)^{8} (-4y)^3 = 165 \cdot (3x)^{8} \cdot (-64y^3) = 165 \cdot 6561x^{8} \cdot (-64y^3) = -69672960x^{8}y^3
  5. (114)(3x)7(−4y)4=330⋅(3x)7⋅(256y4)=330⋅2187x7⋅(256y4)=184492800x7y4{11 \choose 4} (3x)^{7} (-4y)^4 = 330 \cdot (3x)^{7} \cdot (256y^4) = 330 \cdot 2187x^{7} \cdot (256y^4) = 184492800x^{7}y^4
  6. (115)(3x)6(−4y)5=462⋅(3x)6⋅(−1024y5)=462⋅729x6⋅(−1024y5)=−344821760x6y5{11 \choose 5} (3x)^{6} (-4y)^5 = 462 \cdot (3x)^{6} \cdot (-1024y^5) = 462 \cdot 729x^{6} \cdot (-1024y^5) = -344821760x^{6}y^5
  7. (116)(3x)5(−4y)6=462⋅(3x)5⋅(4096y6)=462⋅243x5⋅(4096y6)=460602368x5y6{11 \choose 6} (3x)^{5} (-4y)^6 = 462 \cdot (3x)^{5} \cdot (4096y^6) = 462 \cdot 243x^{5} \cdot (4096y^6) = 460602368x^{5}y^6
  8. (117)(3x)4(−4y)7=330⋅(3x)4⋅(−16384y7)=330⋅81x4⋅(−16384y7)=−440401920x4y7{11 \choose 7} (3x)^{4} (-4y)^7 = 330 \cdot (3x)^{4} \cdot (-16384y^7) = 330 \cdot 81x^{4} \cdot (-16384y^7) = -440401920x^{4}y^7
  9. (118)(3x)3(−4y)8=165⋅(3x)3⋅(65536y8)=165⋅27x3⋅(65536y8)=292669440x3y8{11 \choose 8} (3x)^{3} (-4y)^8 = 165 \cdot (3x)^{3} \cdot (65536y^8) = 165 \cdot 27x^{3} \cdot (65536y^8) = 292669440x^{3}y^8
  10. (119)(3x)2(−4y)9=55⋅(3x)2⋅(−262144y9)=55⋅9x2⋅(−262144y9)=−129766400x2y9{11 \choose 9} (3x)^{2} (-4y)^9 = 55 \cdot (3x)^{2} \cdot (-262144y^9) = 55 \cdot 9x^{2} \cdot (-262144y^9) = -129766400x^{2}y^9
  11. (1110)(3x)1(−4y)10=11⋅(3x)1⋅(1048576y10)=11⋅3x⋅(1048576y10)=34686432xy10{11 \choose 10} (3x)^{1} (-4y)^{10} = 11 \cdot (3x)^{1} \cdot (1048576y^{10}) = 11 \cdot 3x \cdot (1048576y^{10}) = 34686432xy^{10}
  12. (1111)(3x)0(−4y)11=1⋅1⋅(−4194304y11)=−4194304y11{11 \choose 11} (3x)^{0} (-4y)^{11} = 1 \cdot 1 \cdot (-4194304y^{11}) = -4194304y^{11}

Combining these terms, the expansion of (3x−4y)11(3x - 4y)^{11} is:

(3x−4y)11=177147x11−2598156x10y+17372160x9y2−69672960x8y3+184492800x7y4−344821760x6y5+460602368x5y6−440401920x4y7+292669440x3y8−129766400x2y9+34686432xy10−4194304y11(3x - 4y)^{11} = 177147x^{11} - 2598156x^{10}y + 17372160x^{9}y^2 - 69672960x^{8}y^3 + 184492800x^{7}y^4 - 344821760x^{6}y^5 + 460602368x^{5}y^6 - 440401920x^{4}y^7 + 292669440x^{3}y^8 - 129766400x^{2}y^9 + 34686432xy^{10} - 4194304y^{11}

Observations

Notice the alternating signs in the expansion due to the (−4y)(-4y) term. Also, observe how the powers of xx decrease from 11 to 0, while the powers of yy increase from 0 to 11.

Tips for Accuracy

  • Double-check coefficients: Ensure the coefficients from Pascal's Triangle are correct.
  • Pay attention to signs: Keep track of the signs, especially when dealing with negative terms.
  • Simplify carefully: Simplify each term meticulously to avoid errors.

Conclusion

Using Pascal's Triangle to expand binomials like (3x−4y)11(3x - 4y)^{11} simplifies what could be a complex process. By understanding the structure of Pascal's Triangle and applying the binomial theorem, you can efficiently expand binomials to any power. Remember to take your time, double-check your work, and simplify each term carefully for accurate results. With practice, expanding binomials using Pascal's Triangle will become second nature.

For more information on Pascal's Triangle and the Binomial Theorem, visit Khan Academy's resource on the Binomial Theorem.