Understanding the Binomial Expansion and Term Sums

Let's consider the binomial expansion of $\left(x + a\right)^n$. According to the binomial theorem, the expansion is given by:

$\left(x + a\right)^n = \binom{n}{0} x^n a^0 + \binom{n}{1} x^{n-1} a^1 + \binom{n}{2} x^{n-2} a^2 + \binom{n}{3} x^{n-3} a^3 + \dots + \binom{n}{n} x^0 a^n$

The terms in this expansion are indexed starting from 0. The odd terms (1st, 3rd, 5th, etc.) correspond to the terms where the index of the binomial coefficient is even (0, 2, 4, ...). The even terms (2nd, 4th, 6th, etc.) correspond to the terms where the index of the binomial coefficient is odd (1, 3, 5, ...).

Identifying the Sum of Odd and Even Terms

We are given that the sum of odd terms is $P$. The odd terms are the 1st, 3rd, 5th terms and so on. These terms have coefficients $\binom{n}{0}, \binom{n}{2}, \binom{n}{4}$, etc.

So, $P = \binom{n}{0} x^n a^0 + \binom{n}{2} x^{n-2} a^2 + \binom{n}{4} x^{n-4} a^4 + \dots$

We are also given that the sum of even terms is $Q$. The even terms are the 2nd, 4th, 6th terms and so on. These terms have coefficients $\binom{n}{1}, \binom{n}{3}, \binom{n}{5}$, etc.

So, $Q = \binom{n}{1} x^{n-1} a^1 + \binom{n}{3} x^{n-3} a^3 + \binom{n}{5} x^{n-5} a^5 + \dots$

The total binomial expansion is the sum of the odd and even terms:

$\left(x + a\right)^n = P + Q$

Exploring the Expansion of (x - a)n

Now consider the binomial expansion of $\left(x - a\right)^n$:

$\left(x - a\right)^n = \binom{n}{0} x^n \left(-a\right)^0 + \binom{n}{1} x^{n-1} \left(-a\right)^1 + \binom{n}{2} x^{n-2} \left(-a\right)^2 + \binom{n}{3} x^{n-3} \left(-a\right)^3 + \dots$

$= \binom{n}{0} x^n a^0 - \binom{n}{1} x^{n-1} a^1 + \binom{n}{2} x^{n-2} a^2 - \binom{n}{3} x^{n-3} a^3 + \dots$

Notice that the terms with even powers of $a$ (and thus even indices of the binomial coefficient) are the same as the terms in $P$. The terms with odd powers of $a$ (and thus odd indices of the binomial coefficient) are the same as the terms in $Q$, but with a negative sign.

So, $\left(x - a\right)^n = \left(\binom{n}{0} x^n a^0 + \binom{n}{2} x^{n-2} a^2 + \dots\right) - \left(\binom{n}{1} x^{n-1} a^1 + \binom{n}{3} x^{n-3} a^3 + \dots\right)$

$\left(x - a\right)^n = P - Q$

Calculating the Required Expression

We need to find the value of $\left(x + a\right)^{2n} - \left(x - a\right)^{2n}$.

We can rewrite this using the property $\left(b^m\right)^k = b^{mk}$: $\left(x + a\right)^{2n} - \left(x - a\right)^{2n} = \left(\left(x + a\right)^n\right)^2 - \left(\left(x - a\right)^n\right)^2$

This is in the form of a difference of squares, $A^2 - B^2 = (A - B)(A + B)$, where $A = \left(x + a\right)^n$ and $B = \left(x - a\right)^n$.

Substitute the expressions for $A$ and $B$ using $P$ and $Q$:

• $A = \left(x + a\right)^n = P + Q$
• $B = \left(x - a\right)^n = P - Q$

Now calculate $A-B$ and $A+B$:

• $A - B = (P + Q) - (P - Q) = P + Q - P + Q = 2Q$
• $A + B = (P + Q) + (P - Q) = P + Q + P - Q = 2P$

Finally, multiply these results:

$\left(\left(x + a\right)^n\right)^2 - \left(\left(x - a\right)^n\right)^2 = (A - B)(A + B) = (2Q)(2P) = 4PQ$

Conclusion

The value of $\left(x + a\right)^{2n} - \left(x - a\right)^{2n}$ is $4PQ$. This result directly follows from understanding the structure of the binomial expansion and how the sum of odd terms and the sum of even terms relate to $\left(x+a\right)^n$ and $\left(x-a\right)^n$. This demonstrates a useful application of binomial properties.