Question

use pascals triangle to complete the expansion of (x + y)^6. x^6 + 6x^5y + x^4y^2 + 20x^3y^3 + 15x^2y^4 + 6xy^5 + y^6

Explanation:

Step1: Recall Pascal's Triangle for \(n = 6\)

Pascal's Triangle rows correspond to the binomial coefficients of \((x + y)^n\). The row for \(n = 6\) (starting from \(n = 0\)) is: \(1, 6, 15, 20, 15, 6, 1\).

Step2: Identify the coefficient for \(x^4y^2\)

In the expansion of \((x + y)^6\), the term \(x^4y^2\) has a binomial coefficient corresponding to the third element (index 2, 0 - based) in the row for \(n = 6\), which is \(15\)? Wait, no, wait. Wait, the row for \(n = 0\): \(1\); \(n = 1\): \(1, 1\); \(n = 2\): \(1, 2, 1\); \(n = 3\): \(1, 3, 3, 1\); \(n = 4\): \(1, 4, 6, 4, 1\); \(n = 5\): \(1, 5, 10, 10, 5, 1\); \(n = 6\): \(1, 6, 15, 20, 15, 6, 1\). Wait, the expansion of \((x + y)^6\) is \(x^6 + 6x^5y + 15x^4y^2 + 20x^3y^3 + 15x^2y^4 + 6xy^5 + y^6\). Wait, but let's check the pattern. The coefficients for \((x + y)^n\) are given by \(\binom{n}{k}\) where \(k\) is the power of \(y\). For the term \(x^{6 - k}y^k\), the coefficient is \(\binom{6}{k}\). For \(k = 2\) (since \(x^4y^2\) means \(6 - 2 = 4\) for \(x\), \(k = 2\) for \(y\)), \(\binom{6}{2}=\frac{6!}{2!(6 - 2)!}=\frac{6\times5}{2\times1}=15\)? Wait, no, wait the given expansion has \(20x^3y^3\) which is \(\binom{6}{3}=20\), correct. Then for \(x^4y^2\), \(k = 2\), so \(\binom{6}{2}=\frac{6!}{2!4!}=\frac{6\times5}{2\times1}=15\)? But wait the next term after \(6x^5y\) (which is \(k = 1\), \(\binom{6}{1}=6\)) should be \(k = 2\), \(\binom{6}{2}=15\)? Wait but the user's given expansion has \(20x^3y^3\) (which is \(k = 3\), \(\binom{6}{3}=20\)), then \(k = 4\) is \(15x^2y^4\) (\(\binom{6}{4}=15\)), \(k = 5\) is \(6xy^5\) (\(\binom{6}{5}=6\)), \(k = 6\) is \(y^6\) (\(\binom{6}{6}=1\)). Wait, but the term \(x^4y^2\) is \(k = 2\), so \(\binom{6}{2}=15\)? Wait no, wait I think I made a mistake. Wait the row for \(n = 6\) is \(1, 6, 15, 20, 15, 6, 1\). So the coefficients are in order: for \(x^6y^0\): 1, \(x^5y^1\): 6, \(x^4y^2\): 15, \(x^3y^3\): 20, \(x^2y^4\): 15, \(x^1y^5\): 6, \(x^0y^6\): 1. Wait, but the user's problem shows \(x^6 + 6x^5y + \square x^4y^2 + 20x^3y^3 + 15x^2y^4 + 6xy^5 + y^6\). So the missing coefficient is the coefficient for \(x^4y^2\), which is 15? Wait no, wait 6 (for \(x^5y\)) then next is 15? But wait 6, then 15, then 20? Wait no, the row is 1, 6, 15, 20, 15, 6, 1. So the third term (after 1 and 6) is 15. So the coefficient for \(x^4y^2\) is 15? Wait but let's check with binomial theorem. The binomial theorem states \((x + y)^n=\sum_{k = 0}^{n}\binom{n}{k}x^{n - k}y^k\). For \(n = 6\), \(k = 2\): \(\binom{6}{2}=\frac{6!}{2!4!}=\frac{6\times5}{2\times1}=15\). So the coefficient is 15. Wait but the user's problem has \(20x^3y^3\) which is \(k = 3\), \(\binom{6}{3}=20\), correct. So the missing coefficient is 15? Wait no, wait I think I messed up. Wait the row for \(n = 6\) is:

Row 0: 1

Row 1: 1, 1

Row 2: 1, 2, 1

Row 3: 1, 3, 3, 1

Row 4: 1, 4, 6, 4, 1

Row 5: 1, 5, 10, 10, 5, 1

Row 6: 1, 6, 15, 20, 15, 6, 1

Yes, so the coefficients for \((x + y)^6\) are 1 (x^6y^0), 6 (x^5y^1), 15 (x^4y^2), 20 (x^3y^3), 15 (x^2y^4), 6 (x^1y^5), 1 (x^0y^6). So the missing coefficient is 15? Wait but the user's problem shows \(x^6 + 6x^5y + \square x^4y^2 + 20x^3y^3 + 15x^2y^4 + 6xy^5 + y^6\). So the square should be 15? Wait no, wait 6 (k=1), then k=2: 15, k=3:20, k=4:15, k=5:6, k=6:1. So yes, the coefficient for \(x^4y^2\) is 15. Wait but wait, maybe I made a mistake. Wait let's calculate \(\binom{6}{2}\):

\(\binom{6}{2}=\frac{6!}{2!(6 - 2)!}=\frac{6\times5\times4!}{2\times1\times4!}=\frac{30}{2}=15\). Correct. So the missing coefficient is 15.

Answer:

15