Question

use pascals triangle to complete the expansion of ((v - w)^7).
(v^7 - 7v^6w + 21v^5w^2 - 35v^4w^3 + square - 21v^2w^5 + 7vw^6 - w^7)

Explanation:

Step1: Recall Pascal's Triangle for \(n = 7\)

Pascal's Triangle for the 7th power (since the exponent is 7) has coefficients: 1, 7, 21, 35, 35, 21, 7, 1.

Step2: Analyze the pattern of the expansion

The expansion of \((a + b)^n\) follows the pattern where the exponents of \(a\) decrease from \(n\) to 0 and exponents of \(b\) increase from 0 to \(n\), with coefficients from Pascal's Triangle. For \((v - w)^7\), it's equivalent to \((v + (-w))^7\), so the signs alternate starting with positive.
Looking at the given terms:
- The first term: \(v^7\) (coefficient 1, \(v^7w^0\))
- Second term: \(-7v^6w\) (coefficient 7, \(v^6w^1\), sign negative because \((-w)^1=-w\))
- Third term: \(21v^5w^2\) (coefficient 21, \(v^5w^2\), sign positive because \((-w)^2 = w^2\))
- Fourth term: \(-35v^4w^3\) (coefficient 35, \(v^4w^3\), sign negative because \((-w)^3=-w^3\))
- Fifth term: We need to find this. The exponent of \(v\) should be \(7 - 4=3\) (wait, no, exponent of \(v\) in fourth term is 4, so next term (fifth) should have exponent of \(v\) as \(4 - 1 = 3\)? Wait, no, let's check the exponents of \(v\): 7, 6, 5, 4, so next is 3; exponents of \(w\): 0, 1, 2, 3, so next is 4. The coefficient for the fifth term (since the fourth term has coefficient 35, the next coefficient in Pascal's Triangle for \(n = 7\) is also 35 (since the 7th row is 1,7,21,35,35,21,7,1). Now, the sign: \((-w)^4=w^4\), so the sign is positive. So the term should be \(35v^3w^4\)? Wait, no, wait the sign: Wait, let's re - check the sign pattern.
Wait, \((v - w)^7=\sum_{k = 0}^{7}\binom{7}{k}v^{7 - k}(-w)^{k}\)
For \(k = 0\): \(\binom{7}{0}v^{7}(-w)^{0}=v^{7}\)
\(k = 1\): \(\binom{7}{1}v^{6}(-w)^{1}=-7v^{6}w\)
\(k = 2\): \(\binom{7}{2}v^{5}(-w)^{2}=21v^{5}w^{2}\)
\(k = 3\): \(\binom{7}{3}v^{4}(-w)^{3}=-35v^{4}w^{3}\)
\(k = 4\): \(\binom{7}{4}v^{3}(-w)^{4}\)
Since \(\binom{7}{4}=\binom{7}{3}=35\) (because \(\binom{n}{k}=\binom{n}{n - k}\)), and \((-w)^{4}=w^{4}\), so this term is \(35v^{3}w^{4}\)
Wait, but let's check the given terms after the fourth term: the next term after \(-35v^{4}w^{3}\) is the one we need to find, then after that is \(-21v^{2}w^{5}\). Wait, no, the exponent of \(v\) in the term after \(-35v^{4}w^{3}\): exponent of \(v\) should be \(4-1 = 3\), exponent of \(w\) should be \(3 + 1=4\). The coefficient is 35 (from Pascal's Triangle: 1,7,21,35,35,21,7,1). Now, the sign: \((-w)^4=w^4\), so the sign is positive. But wait, the term after our missing term is \(-21v^{2}w^{5}\). Let's check the exponent of \(v\) in \(-21v^{2}w^{5}\): exponent of \(v\) is 2, so before that (our missing term) exponent of \(v\) is 3, exponent of \(w\) is 4. The coefficient is 35, and the sign: since \((-w)^4 = w^4\), the term is positive. But wait, let's check the pattern of signs. The signs are: +, -, +, -, +, -, +, -. Wait, first term: +, second: -, third: +, fourth: -, fifth: +, sixth: -, seventh: +, eighth: -. Let's check the given terms:
- \(v^7\): + (correct, \(k = 0\), \((-w)^0=1\), sign +)
- \(-7v^6w\): - (correct, \(k = 1\), \((-w)^1=-w\), sign -)
- \(21v^5w^2\): + (correct, \(k = 2\), \((-w)^2 = w^2\), sign +)
- \(-35v^4w^3\): - (correct, \(k = 3\), \((-w)^3=-w^3\), sign -)
- Next term (fifth, \(k = 4\)): \((-w)^4=w^4\), so sign +, coefficient 35, \(v^{7 - 4}w^{4}=v^3w^4\). So the term is \(35v^3w^4\)? Wait, but the term after that is \(-21v^2w^5\). Let's check the coefficient for \(k = 5\): \(\binom{7}{5}=21\), \((-w)^5=-w^5\), so term is \(-21v^{2}w^5\) (which matches). Then \(k = 6\): \(\binom{7}{6}=7\), \((-w)^6 = w^6\), term \(7vw^6\) (matches), \(k = 7\): \(\binom{7}{7}=1\), \((-w)^7=-w^7\) (matches). Wait, but in the given expansion, the missing term is between \(-35v^4w^3\) and \(-21v^2w^5\). Wait, exponent of \(v\) in \(-35v^4w^3\) is 4, in the missing term exponent of \(v\) is 3, in \(-21v^2w^5\) exponent of \(v\) is 2. Exponent of \(w\) in \(-35v^4w^3\) is 3, in missing term is 4, in \(-21v^2w^5\) is 5. The coefficient for \(k = 4\) is 35, and the sign: since \(k = 4\), \((-w)^4=w^4\), so the sign is positive. But wait, the term after our missing term is \(-21v^2w^5\) (which is \(k = 5\), sign -). So our missing term ( \(k = 4\)) should have sign +. But let's check the coefficients again. The 7th row of Pascal's Triangle is 1,7,21,35,35,21,7,1. So the coefficients for the terms (in order) are 1,7,21,35,35,21,7,1. So the fifth term (counting from 0) has coefficient 35. So the term is \(+35v^{3}w^{4}\)? But wait, the given expansion has after \(-35v^4w^3\) a blank, then \(-21v^2w^5\). Wait, maybe I made a mistake in the exponent of \(v\). Wait, \((v - w)^7=\sum_{i = 0}^{7}\binom{7}{i}v^{7 - i}(-w)^{i}\). So for \(i = 0\): \(i = 0\), term: \(\binom{7}{0}v^{7}(-w)^0=v^7\)
\(i = 1\): \(\binom{7}{1}v^{6}(-w)^1=-7v^6w\)
\(i = 2\): \(\binom{7}{2}v^{5}(-w)^2=21v^5w^2\)
\(i = 3\): \(\binom{7}{3}v^{4}(-w)^3=-35v^4w^3\)
\(i = 4\): \(\binom{7}{4}v^{3}(-w)^4=35v^3w^4\) (because \((-w)^4 = w^4\), so positive sign)
\(i = 5\): \(\binom{7}{5}v^{2}(-w)^5=-21v^2w^5\) (because \((-w)^5=-w^5\), negative sign)
\(i = 6\): \(\binom{7}{6}v^{1}(-w)^6=7vw^6\) (because \((-w)^6 = w^6\), positive sign)
\(i = 7\): \(\binom{7}{7}v^{0}(-w)^7=-w^7\) (because \((-w)^7=-w^7\), negative sign)
Ah! Wait, the term with \(i = 4\) is \(35v^{3}w^{4}\), but the given expansion after the blank is \(-21v^2w^5\) (which is \(i = 5\)). So the blank should be \(35v^3w^4\)? But wait, the sign: let's check the sign again. For \(i = 4\), \((-w)^4 = w^4\), so the term is positive, so \(+35v^3w^4\). But the given expansion has after \(-35v^4w^3\) a blank, then \(-21v^2w^5\). So the blank is \(35v^3w^4\)? Wait, but let's check the exponents of \(v\) and \(w\) again. In the term \(-35v^4w^3\), exponent of \(v\) is 4, exponent of \(w\) is 3. In the next term (our blank), exponent of \(v\) should be \(4 - 1=3\), exponent of \(w\) should be \(3 + 1 = 4\). So \(v^3w^4\), coefficient 35, sign positive (because \((-w)^4=w^4\)). So the term is \(35v^3w^4\). Wait, but let's check the coefficient sequence. The coefficients are 1,7,21,35,35,21,7,1. So after 35 (the fourth term's coefficient), the next coefficient is 35 (the fifth term's coefficient). So yes, the missing term is \(35v^3w^4\). But wait, the sign: the fourth term is \(-35v^4w^3\) (sign -), the fifth term (our blank) should have sign +, then the sixth term is \(-21v^2w^5\) (sign -), which matches the alternating sign pattern (+, -, +, -, +, -, +, -). So the missing term is \(35v^3w^4\).

Answer:

\(35v^{3}w^{4}\)