Question

If the number of terms in the expansion \((2 x+y)^{n}-(2 x-y)^{n}\) is 8 , then the value of \(n\) is (where \(\mathrm{n}\) is odd (1) 17 (2) 19 (3) 15 (4) 13

Short answer

Answer: The possible values for n are 13 and 17.

Step-by-step solution

Find the general term for the expansion

Using the binomial theorem, the general term for \((2x+y)^n\) and \((2x-y)^n\) can be written as: \(T_k = \binom{n}{k}(2x)^{n-k}(y)^k\) and \(T'_k=\binom{n}{k}(2x)^{n-k}(-y)^k\) for the expansions \((2x+y)^n\) and \((2x-y)^n\) respectively, where \(k\) ranges from \(0\) to \(n\).

Find the condition when the difference of two terms cancel out

We know that when the difference of the two expansions has 8 terms, some terms from both expansions should cancel out each other. The condition for cancellation is when the terms from both expansions have the same coefficients but opposite signs. In other words, for cancellation of \(T_k\) and \(T'_k\), we should have: \(\binom{n}{k}(2x)^{n-k}(y)^k = -\binom{n}{k}(2x)^{n-k}(-y)^k\)

Simplify the condition and find the value of k

The above condition simplifies to: \((y)^k = -(-y)^k\), which means one of them should be positive, and the other should be negative. Since the power of \(y\) is even in \((-y)^k\), the term would be positive, so \((y)^k\) should be negative. This is possible only when\\ \(k\) is odd: \(k = 2m+1\), where \(m\) is an integer.

Find the condition for the number of terms in the expansions after cancellations

Since there are \((n+1)\) terms in each expansion and after the cancellation, 8 terms remain, the number of terms that cancel out in the expansions should be equal to \(\frac{n+1-8}{2}\) (each term from the expansions cancels out one term from the other expansion). So, we have: \(\frac{n+1-8}{2} = n - 2m\), where \(n\) and \(k\) should be odd integers.

Check for which option the condition is met

Now we can check the given options (1) 17, (2) 19, (3) 15, (4) 13, one by one to find the correct value of odd integer \(n\). (1) If n = 17, then \(m = \frac{17-7}{2} = 5\), which is an integer, so this option is valid. (2) If n = 19, then \(m = \frac{19-7}{2} = 6\), but since we need an odd integer value for \(m\), this option is not valid. (3) If n = 15, then \(m = \frac{15-7}{2} = 4\), but since we need an odd integer value for \(m\), this option is not valid. (4) If n = 13, then \(m = \frac{13-7}{2} = 3\), which is an integer, so this option is valid. So, the value of \(n\) is odd and can take two values in the given options, \(\boxed{(1) 17}\) and \(\boxed{(4) 13}\).

Key concepts

Expansion

The binomial expansion is a powerful algebraic tool for expressing any power of a binomial expression. In simpler terms, it's a way to break down an expression like \((a+b)^n\) into a sum of terms of the form \(\binom{n}{k}a^{n-k}b^k\). This is known as the binomial theorem. It tells us that each term in the expansion is determined by a binomial coefficient, and the powers of each part of the binomial move in opposite directions. For instance, in \((2x+y)^n\), terms expand out as a series with powers of \(2x\) decreasing while those of \(y\) increase.The expansion helps us solve the problem involving finding the number of terms present when simplifying expressions. By determining these terms, we can further analyze how expressions like \((2x+y)^n - (2x-y)^n\) behave.

Odd Integers

Odd integers play a significant role in our exercise because they influence whether terms cancel out when simplifying an equation. Odd integers, like 1, 3, 5, and so on, do not evenly divide by 2. They are crucial when considering sign changes in powers, especially when dealing with terms like \((y)^k\) and \((-y)^k\).In the problem, we identified conditions where only terms with odd powers cancel each other out. This is because odd powers of negative numbers switch the entire term's sign. Therefore, if a given expansion leads to a situation where certain terms should subtract to zero, it is often the odd-powered terms that make this possible.Understanding how odd integers fall into play in expansion problems allows us to predict which terms will remain after others cancel out. This is a core part of finding solutions that precisely match conditions like those in this exercise.

Coefficient Cancellation

Coefficient cancellation is when two terms in a binomial expansion cancel each other out due to having equal magnitudes with opposite signs. This is important in simplifying expressions, such as \((2x+y)^n-(2x-y)^n\), as it explains how terms disappear from the final expression.Using the general term in the binomial expansion, \(T_k - T'_k = \binom{n}{k}(2x)^{n-k}(y)^k + \binom{n}{k}(2x)^{n-k}(-y)^k\), we find that if continuing cancellation leaves only the chosen few with specific properties (e.g., odd k), many terms do not appear in the expression's expansion.

• For example, the terms \((y)^k\) and \((-y)^k\) cancel each other when \(k\) is odd.
• Their cancellations shape the structure of the resulting expression efficiently.

Calculating different configurations could determine the cases where terms reduce to zero, significantly aiding simplification and reaching the correct answers.

General Term

The general term provides a crucial framework for identifying specific terms within a binomial expansion. It is expressed using the formula: \[ T_k = \binom{n}{k}(2x)^{n-k}(y)^k \] for \((2x+y)^n\) and \[ T'_k = \binom{n}{k}(2x)^{n-k}(-y)^k \] for \((2x-y)^n\). In these expansions, the term's coefficient, \(\binom{n}{k}\) (the binomial coefficient), tells us how many different ways the constants \(a\) and \(b\) appear across the different groups of each term.The general term is critical for identifying the structure and finding specific values in an expanded expression. In our problem, this is central to determining when terms cancel (using odd values for \(k\) for example) and in adjusting expressions to achieve the required number of surviving terms (such as 8 non-cancelled terms). Understanding how to manipulate the general term aids in situating specific values of \(n\) that satisfy the structured conditions of chosen exercises.