in the expansion of (x+a)^n if the sum of odd terms be P & sum of even terms be q prove that

To prove the relationship between the sums of odd and even terms in the expansion of $(x + a)^n$, we can use the binomial theorem and some algebraic manipulations.

1. Break Down the Problem

We need to express the sums of the odd and even terms in the binomial expansion of $(x + a)^n$ and analyze their relationship.

2. Relevant Concepts

The binomial expansion states that: $(x + a)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} a^k$ Where $\binom{n}{k}$ is the binomial coefficient.

3. Analysis and Detail

• Odd Terms (where k is odd): The terms for odd $k$ can be represented as: $\text{Sum of odd terms} \, P = \sum_{\text{odd } k} \binom{n}{k} x^{n-k} a^k$

• Even Terms (where k is even): The terms for even $k$ can be represented as: $\text{Sum of even terms} \, Q = \sum_{\text{even } k} \binom{n}{k} x^{n-k} a^k$

• The total sum of the expansion can be represented as: $S = P + Q$

• The sums for specific cases can be formulated using the properties of the binomial expansion. We can substitute $x = 1$ and $x = -1$ to isolate the sums.

4. Isolating P and Q

1. Set $x = 1$: $S(1) = (1 + a)^n = P + Q$

2. Set $x = -1$: $S(-1) = (-1 + a)^n = P - Q$

3. Now we have two equations: $P + Q = (1 + a)^n$ $P - Q = (-1 + a)^n$

4. By adding these two equations, we can find $P$: $2P = (1 + a)^n + (-1 + a)^n$ Thus, $P = \frac{(1 + a)^n + (-1 + a)^n}{2}$

5. By subtracting the second equation from the first, we find $Q$: $2Q = (1 + a)^n - (-1 + a)^n$ Thus, $Q = \frac{(1 + a)^n - (-1 + a)^n}{2}$

5. Verify and Summarize

• The sums of the odd and even terms have been expressed in terms of the powers of $(1 + a)$ and $(-1 + a)$.
• Therefore, when we summarize, we can see how $P$ and $Q$ relate to each other using these forms.

Final Answer

$P + Q = (1 + a)^n, \quad P - Q = (-1 + a)^n$ Thus, the relationship between the sums of the odd and even terms in the expansion of $(x + a)^n$ is valid based on the derived expressions for $P$ and $Q$.