Let α, β, γ and δ be the coefficients of x^7, x^5, x^3 and x respectively in the expansion of (x + √

Q: Wait, why do the odd powers of $B$ cancel out?

When you expand $(A+B)^n$ and $(A-B)^n$, the terms in the second expansion have alternating signs. Specifically, the $k$-th term in $(A-B)^n$ is $(-1)^k \binom{n}{k} A^{n-k} B^k$. When you add them, terms with odd $k$ have opposite signs and cancel out, while terms with even $k$ add up, resulting in $2 \sum \binom{n}{k} A^{n-k} B^k$ for even $k$.

Q: How do I know when to stop the expansion?

The expansion of $(A+B)^5$ goes up to $k=5$. However, our identity for $(A+B)^n + (A-B)^n$ only includes even $k$. Since $k$ must be $\le 5$, the even values are $0, 2, 4$. The next even value would be $6$, which is greater than $n=5$, so it doesn't exist.

Q: Is there a faster way than expanding everything?

You could potentially use derivatives or specific values of $x$ to find the coefficients, but for a polynomial of degree 7, direct expansion is often the most reliable and straightforward method. It minimizes the risk of missing a term and is quite manageable for $n=5$.

Question

Let $\alpha, \beta, \gamma$ and $\delta$ be the coefficients of $x^{7}$, $x^{5}$, $x^{3}$ and $x$ respectively in the expansion of $(x+\sqrt{x^{3}-1})^{5}+(x-\sqrt{x^{3}-1})^{5}$, $x>1$. If u and v satisfy the equations $\alpha u+\beta v=18$ and $\gamma u+\delta v=20$, then $u+v$ equals:

Options:

(A) 5

(B) 3

(C) 4

(D) 8

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before this seemingly monstrous expression: $(x+\sqrt{x^3-1})^5 + (x-\sqrt{x^3-1})^5$. It looks like a nightmare of algebra, but in mathematics, complexity is often just a mask for symmetry.

Notice the structure. We have $(A+B)^5 + (A-B)^5$, where $A=x$ and $B=\sqrt{x^3-1}$.

When you expand these two binomials, the terms with odd powers of $B$ will have opposite signs and vanish into thin air. You are left with only the even powers of $B$.

We are essentially calculating:

$$2 \left[ \binom{5}{0} A^5 + \binom{5}{2} A^3 B^2 + \binom{5}{4} A B^4 \right]$$

### The Heavy Lifting

Now, let us substitute our values back in. With $A=x$ and $B=\sqrt{x^3-1}$, the expression becomes:

$$2 \left[ \binom{5}{0} x^5 + \binom{5}{2} x^3 (\sqrt{x^3-1})^2 + \binom{5}{4} x (\sqrt{x^3-1})^4 \right]$$

Calculating the coefficients is straightforward: $\binom{5}{0}=1$, $\binom{5}{2}=10$, and $\binom{5}{4}=5$.

When we square $\sqrt{x^3-1}$, we get $x^3-1$. When we raise it to the power of 4, we get $(x^3-1)^2$, which is $x^6 - 2x^3 + 1$.

Our expression now looks like:

$$2 \left[ x^5 + 10x^3(x^3-1) + 5x(x^6 - 2x^3 + 1) \right]$$

Expanding this gives:

$$2 \left[ x^5 + 10x^6 - 10x^3 + 5x^7 - 10x^4 + 5x \right]$$

Multiplying by 2, we obtain the final polynomial:

$$10x^7 + 20x^6 + 2x^5 - 20x^4 - 20x^3 + 10x$$

### The Coefficient Hunt

We have our polynomial! Now, we just need to extract the coefficients as defined in the problem.

The coefficient of $x^7$ is $\alpha = 10$. The coefficient of $x^5$ is $\beta = 2$. The coefficient of $x^3$ is $\gamma = -20$. The coefficient of $x$ is $\delta = 10$.

### The Final Resolution

We are left with a simple system of linear equations: $10u + 2v = 18$ and $-20u + 10v = 20$.

Simplifying these, we get $5u + v = 9$ and $-2u + v = 2$.

Subtracting the second equation from the first, we get $7u = 7$, so $u=1$.

Substituting $u=1$ into $5u+v=9$, we get $5+v=9$, so $v=4$.

The final answer, $u+v$, is **5**.

Select Answer:

Visualized Solution

Identifying the Structure

The Binomial Identity

Substituting $A$ and $B$

Evaluating Coefficients

Expanding the Terms

Simplifying the Brackets

Final Polynomial Form

Identifying $α, β, γ, δ$

Setting up Equations

Simplifying the Equations

Solving for $u$

Solving for $v$

Final Answer

The Sigma Insight: Binomial Expansion for Positive Integral Index

Analyzing the Setup

The Heavy Lifting

The Coefficient Hunt

The Final Resolution

Similar Questions

If $α$ and $β$ be the coefficients of $x^{4}$ and $x^{2}$ respectively in the expansion of $(x + x^{2} - 1)^{6} + (x - x^{2} - 1)^{6}$ , then:

If in the expansion of $(1 + x)^{p} (1 - x)^{q}$ the coefficients of $x$ and $x^{2}$ are 1 and -2, respectively, then $p^{2} + q^{2}$ is equal to:

If $α$ and $β$ be the coefficient of $x^{4}$ and $x^{2}$ respectively in the expansion of $(x + x^{2} - 1)^{6} + (x - x^{2} - 1)^{6}$ , then :

If the coefficients of $x$ and $x^{2}$ in $(1 + x)^{p} (1 - x)^{q}$ are 4 and $- 5$ respectively, then $2 p + 3 q$ is equal to

Let $n \geq 5$ be an integer. If $9^{n} - 8 n - 1 = 64 α$ and $6^{n} - 5 n - 1 = 25 β$ , then $α - β$ is equal to:

Let the coefficient of $x^{r}$ in the expansion of $(x + 3)^{n - 1} + (x + 3)^{n - 2} (x + 2) + (x + 3)^{n - 3} (x + 2)^{2} + \dots + (x + 2)^{n - 1}$ be $α_{r}$ . If $\sum_{r = 0}^{n} α_{r} = β^{n} - γ^{n}, β, γ \in N$ , then the value of $β^{2} + γ^{2}$ equals _______.

In the expansion of $(1 + x) (1 - x^{2}) (1 + \frac{3}{x} + \frac{3}{x ^{2}} + \frac{1}{x ^{3}})^{5}, x \neq = 0$ , the sum of the coefficient of $x^{3}$ and $x^{- 13}$ is equal to _______.

The sum of the co-efficients of all odd degree terms in the expansion of $(x + x^{3} - 1)^{5} + (x - x^{3} - 1)^{5}, (x > 1)$ is :

The sum of the co-efficients of all even degree terms in $x$ in the expansion of $(x + x^{3} - 1)^{6} + (x - x^{3} - 1)^{6}, (x > 1)$ is equal to :

If in the expansion of $(1 + x)^{m} (1 - x)^{n}$ , the coefficients of $x$ and $x^{2}$ are 3 and $- 6$ respectively, then $m$ is

Select Answer:

Visualized Solution

Identifying the Structure

The Binomial Identity

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Evaluating Coefficients

Expanding the Terms

Simplifying the Brackets

Final Polynomial Form

Identifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α,β,γ,δ

Setting up Equations

Simplifying the Equations

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } u

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v

Final Answer

The Sigma Insight: Binomial Expansion for Positive Integral Index

Analyzing the Setup

The Heavy Lifting

The Coefficient Hunt

The Final Resolution

Similar Questions

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