Problem 7
Question
The coefficients of three consecutive terms in the expansion of \((1+a)^{n}\) are in the ratio \(1: 7: 42\). Find \(n\).
Step-by-Step Solution
Verified Answer
n = 55
1Step 1: Set up the ratio equations
In the expansion of \((1+a)^n\), the coefficient of \(a^r\) is \(\binom{n}{r}\). Three consecutive coefficients are in the ratio \(1:7:42\):
\(\binom{n}{r} : \binom{n}{r+1} : \binom{n}{r+2} = 1 : 7 : 42\)
\(\binom{n}{r} : \binom{n}{r+1} : \binom{n}{r+2} = 1 : 7 : 42\)
2Step 2: Use the ratio of consecutive binomial coefficients
\(\frac{\binom{n}{r+1}}{\binom{n}{r}} = \frac{n-r}{r+1} = 7 \quad \Rightarrow \quad n - r = 7r + 7 \quad \cdots (1)\)
\(\frac{\binom{n}{r+2}}{\binom{n}{r+1}} = \frac{n-r-1}{r+2} = 6 \quad \Rightarrow \quad n - r - 1 = 6r + 12 \quad \cdots (2)\)
\(\frac{\binom{n}{r+2}}{\binom{n}{r+1}} = \frac{n-r-1}{r+2} = 6 \quad \Rightarrow \quad n - r - 1 = 6r + 12 \quad \cdots (2)\)
3Step 3: Solve the system
From (1): \(n - r = 7r + 7\).
From (2): \(n - r = 6r + 13\).
Setting equal: \(7r + 7 = 6r + 13 \Rightarrow r = 6\).
Substituting back: \(n = r + 7r + 7 = 6 + 42 + 7 = 55\).
From (2): \(n - r = 6r + 13\).
Setting equal: \(7r + 7 = 6r + 13 \Rightarrow r = 6\).
Substituting back: \(n = r + 7r + 7 = 6 + 42 + 7 = 55\).
4Step 4: Verify
Check: \(\frac{\binom{55}{7}}{\binom{55}{6}} = \frac{49}{7} = 7\) \(\checkmark\)
\(\frac{\binom{55}{8}}{\binom{55}{7}} = \frac{48}{8} = 6\) \(\checkmark\)
Ratio: \(1 : 7 : 7 \times 6 = 1 : 7 : 42\) \(\checkmark\)
Therefore \(n = \boxed{55}\).
\(\frac{\binom{55}{8}}{\binom{55}{7}} = \frac{48}{8} = 6\) \(\checkmark\)
Ratio: \(1 : 7 : 7 \times 6 = 1 : 7 : 42\) \(\checkmark\)
Therefore \(n = \boxed{55}\).
Key Concepts
CoefficientsConsecutive TermsCombination Formula
Coefficients
Coefficients in a binomial expansion play a crucial role as they determine the numerical part of each term. For an expression like \((1+a)^{n}\), each term of the expansion has a coefficient that reflects the number of ways to choose a particular subset of elements. These coefficients are derived from the binomial theorem, represented by binomial coefficients or combinatory numbers. In our example, the coefficients of the three consecutive terms are in a specific ratio: \(1: 7: 42\). This ratio provides a relationship among the terms, allowing us to solve for \(n\) by equating them to binomial coefficients \(\binom{n}{r-1}\), \(\binom{n}{r}\), and \(\binom{n}{r+1}\). These coefficients help us understand how likely each outcome is in the binomial situation, which in turn informs us about the structure and nature of the expansion itself.
Consecutive Terms
Consecutive terms in a binomial expansion mean the terms follow one after the other without skipping any in between. When solving problems involving these terms, it is important to identify the terms sequentially. In the context of the given problem, the consecutive terms are denoted by \(T_r\), \(T_{r+1}\), and \(T_{r+2}\). These terms are determined by their place in the expansion and each term increases by one power of \(a\) as you move from one term to the next. Understanding consecutive terms is essential because the relationships between them often hold the key to solving the problem. In our exercise, recognizing how these terms interact according to their ratios allows us to form equations that help in determining the sequence and the parameter \(n\). This sequential nature is what allows us to apply reasoning about ratios or consecutive number patterns to find solutions.
Combination Formula
The combination formula is a mathematical tool used to calculate the binomial coefficients in binomial expansions. Given as \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\), it shows how many different ways you can choose \(k\) items from \(n\) items without regard to the order of selection. In our problem, the combination formula is employed to find the coefficients of the terms \(\binom{n}{r-1}, \binom{n}{r}, \text{and} \binom{n}{r+1}\). This use of combinations reflects the number of subsets and thus the coefficients within our expansion, integral to solving binomial problems. Utilizing this formula requires a fundamental understanding of factorials, which denote the product of all positive integers up to a certain number. Using the combination formula allows us to relate these coefficients to the given ratio, unlocking the solution by systematically balancing these equations to find our desired \(n\). Learning this can significantly ease understanding of how these calculations contribute to broader mathematical tasks.
Other exercises in this chapter
Problem 6
Find the coeffcient of \(x^{6} y^{3}\) in the expansion of \((x+2 y)^{9}\).
View solution Problem 6
9 th term in the expansion of \(\left(\frac{y}{2}+2 x\right)^{12}\) is (a) \(7920 x^{7} x^{5}\) (b) \(7920 x^{6} y^{6}\) (c) \(7920 x^{8} y^{4}\) (d) \(7816 x^{
View solution Problem 7
Find the number of terms in the expansions of the following. \((2 x-3 y)^{9}\)
View solution Problem 7
If \(A\) and \(B\) are the coefficient of \(x^{n}\) in the expansions of \((1+x)^{2 n}\) and \((1+x)^{2 n-1}\) respectively, then (a) \(A=B\) (b) \(A=2 B\) (c)
View solution