Finding Non-Zero Terms in Binomial Expansion

The problem asks for the number of non-zero terms in the simplified expansion of the sum of two binomial expressions: \(\left(1 + 2\sqrt{3}x\right)^{11} + \left(1 - 2\sqrt{3}x\right)^{11}\).

Let's consider the general form of the sum of two binomial expansions: \((a+b)^n + (a-b)^n\).

Using the binomial theorem, the expansion of \((a+b)^n\) is given by:

\((a+b)^n = \binom{n}{0} a^n b^0 + \binom{n}{1} a^{n-1} b^1 + \binom{n}{2} a^{n-2} b^2 + \dots + \binom{n}{k} a^{n-k} b^k + \dots + \binom{n}{n} a^0 b^n\)

And the expansion of \((a-b)^n\) is given by:

\((a-b)^n = \binom{n}{0} a^n (-b)^0 + \binom{n}{1} a^{n-1} (-b)^1 + \binom{n}{2} a^{n-2} (-b)^2 + \dots + \binom{n}{k} a^{n-k} (-b)^k + \dots + \binom{n}{n} a^0 (-b)^n\)

This can be written as:

\((a-b)^n = \binom{n}{0} a^n b^0 - \binom{n}{1} a^{n-1} b^1 + \binom{n}{2} a^{n-2} b^2 - \dots + (-1)^k \binom{n}{k} a^{n-k} b^k + \dots + (-1)^n \binom{n}{n} a^0 b^n\)

Now, let's add these two expansions:

\((a+b)^n + (a-b)^n = [\binom{n}{0} a^n b^0 + \binom{n}{1} a^{n-1} b^1 + \binom{n}{2} a^{n-2} b^2 + \dots] + [\binom{n}{0} a^n b^0 - \binom{n}{1} a^{n-1} b^1 + \binom{n}{2} a^{n-2} b^2 - \dots]\)

When we add the two expansions, the terms with odd powers of $b$ (where the exponent $k$ is odd) will cancel out, because they appear with opposite signs. The terms with even powers of $b$ (where the exponent $k$ is even) will be added together, effectively doubling their coefficient.

So, the sum is:

\((a+b)^n + (a-b)^n = 2 \left[ \binom{n}{0} a^n b^0 + \binom{n}{2} a^{n-2} b^2 + \binom{n}{4} a^{n-4} b^4 + \dots \right]\)

The terms that remain are those where the power of $b$ (which corresponds to the index $k$ in \(\binom{n}{k}\)) is an even number.

In the given problem, we have \((1 + 2\sqrt{3}x)^{11} + (1 - 2\sqrt{3}x)^{11}\). Here, $a=1$, \(b=2\sqrt{3}x\), and $n=11$.

The simplified expansion will contain terms corresponding to even values of $k$ in the binomial expansion of \((a+b)^n\), multiplied by 2. The general term is \(2 \binom{11}{k} (1)^{11-k} (2\sqrt{3}x)^k\), where $k$ is even.

The possible values for $k$ in the expansion of \((a+b)^{11}\) range from 0 to 11. We need the even values of $k$ within this range. The even integers from 0 to 11 are:

• $k = 0$
• $k = 2$
• $k = 4$
• $k = 6$
• $k = 8$
• $k = 10$

For each of these values of $k$, we get a term in the simplified expansion:

• For $k=0$: \(2 \binom{11}{0} (1)^{11} (2\sqrt{3}x)^0\). This term is \(2 \times 1 \times 1 \times 1 = 2\). This is a non-zero constant term.
• For $k=2$: \(2 \binom{11}{2} (1)^{9} (2\sqrt{3}x)^2 = 2 \binom{11}{2} (12x^2)\). This is a non-zero term involving \(x^2\).
• For $k=4$: \(2 \binom{11}{4} (1)^{7} (2\sqrt{3}x)^4 = 2 \binom{11}{4} (144x^4)\). This is a non-zero term involving \(x^4\).
• For $k=6$: \(2 \binom{11}{6} (1)^{5} (2\sqrt{3}x)^6\). This is a non-zero term involving \(x^6\).
• For $k=8$: \(2 \binom{11}{8} (1)^{3} (2\sqrt{3}x)^8\). This is a non-zero term involving \(x^8\).
• For $k=10$: \(2 \binom{11}{10} (1)^{1} (2\sqrt{3}x)^{10}\). This is a non-zero term involving \(x^{10}\).

Since the variable terms have different powers of $x$ (\(x^0, x^2, x^4, x^6, x^8, x^{10}\)), they are distinct terms and will not combine further. The coefficients \(2 \binom{11}{k} (2\sqrt{3})^k\) for $k = 0, 2, 4, 6, 8, 10$ are all non-zero.

The number of such terms is the count of the even values of $k$ from 0 to 10, which is 6.

There are 6 terms in the simplified expansion, and all of them are non-zero.

Revision Table: Binomial Expansion Concepts

Additional Information on Non-Zero Terms

When expanding binomials of the form \((P + Q)^n\) where $P$ and $Q$ are expressions involving a variable (like $x$), the terms in the expansion are generally distinct based on the power of the variable $x$. For example, in \((1+x)^3 = 1 + 3x + 3x^2 + x^3\), the terms are \(1, 3x, 3x^2, x^3\), which have distinct powers of $x$ (\(x^0, x^1, x^2, x^3\)) and are all non-zero. If any coefficient were zero, that term would disappear.

In the case of \((a+b)^n + (a-b)^n\), where $b$ contains the variable $x$, the simplified expansion results in terms where the power of $x$ is $k$ (specifically $2k$ in the simplified sum formula). As long as the coefficient \(2 \binom{n}{k} a^{n-k}\) and the variable part \(b^k\) are non-zero for the relevant even $k$, the resulting terms will be non-zero and distinct due to the different powers of $x$.

For the expression \(\left(1 + 2\sqrt{3}x\right)^{11} + \left(1 - 2\sqrt{3}x\right)^{11}\), the terms only involve even powers of $x$ (from \(x^0\) up to \(x^{10}\)). Since the coefficients are non-zero, each term with a unique power of $x$ contributes a non-zero term to the total sum. The powers are \(x^0, x^2, x^4, x^6, x^8, x^{10}\), giving a total of 6 unique terms, all of which are non-zero.