Understanding the Algebraic Expansion Problem

The question asks us to find the total number of terms in the expansion of the expression \((1 + 2x + x^2)^5 + (1 + 4y + 4y^2)^5\). To solve this, we first need to simplify the terms within the parentheses and then consider the properties of binomial expansions and how terms combine when expressions are added.

Simplifying the Expressions

Let's look at the expressions inside the parentheses:

• The first part is \((1 + 2x + x^2)\). This is a perfect square trinomial, which can be written as \((1+x)^2\).
• The second part is \((1 + 4y + 4y^2)\). This is also a perfect square trinomial, which can be written as \((1+2y)^2\).

So, the given expression becomes:

\(( (1+x)^2 )^5 + ( (1+2y)^2 )^5\)

Using the exponent rule \((a^m)^n = a^{mn}\), we can simplify this further:

\((1+x)^{2 \times 5} + (1+2y)^{2 \times 5}\)

This simplifies to:

\((1+x)^{10} + (1+2y)^{10}\)

Number of Terms in a Binomial Expansion

The binomial theorem states that the expansion of \((a+b)^n\) has \(n+1\) terms. Let's apply this to each part of our simplified expression:

• The first part is \((1+x)^{10}\). Here, \(n=10\). So, the expansion of \((1+x)^{10}\) will have 10+1 = 11 terms. These terms will be of the form \(\binom{10}{k} 1^{10-k} x^k = \binom{10}{k} x^k\) for \(k=0, 1, 2, \dots, 10\). The powers of \(x\) range from \(x^0\) to \(x^{10}\).
• The second part is \((1+2y)^{10}\). Here also, \(n=10\). So, the expansion of \((1+2y)^{10}\) will have 10+1 = 11 terms. These terms will be of the form \(\binom{10}{j} 1^{10-j} (2y)^j = \binom{10}{j} 2^j y^j\) for \(j=0, 1, 2, \dots, 10\). The powers of \(y\) range from \(y^0\) to \(y^{10}\).

Combining Terms from Two Expansions

Now we need to find the number of terms in the sum of these two expansions: \((1+x)^{10} + (1+2y)^{10}\).

The expansion of \((1+x)^{10}\) consists of terms with different powers of \(x\) (from \(x^0\) to \(x^{10}\)):

The expansion of \((1+2y)^{10}\) consists of terms with different powers of \(y\) (from \(y^0\) to \(y^{10}\)):

Since the variables \(x\) and \(y\) are different, a term involving \(x^k\) (where \(k > 0\)) cannot be combined with a term involving \(y^j\) (where \(j > 0\)). The only terms that could potentially combine are the constant terms (terms with \(x^0\) and \(y^0\)).

Let's look at the terms based on their variables:

• Constant terms: The constant term in \((1+x)^{10}\) is \(\binom{10}{0}x^0 = 1\). The constant term in \((1+2y)^{10}\) is \(\binom{10}{0}(2y)^0 = 1\). When we add the expansions, the constant term will be the sum of these two constant terms: 1 + 1 = 2. This contributes 1 term to the total expansion.
• Terms involving powers of x: These terms come only from the expansion of \((1+x)^{10}\). They are of the form \(\binom{10}{k}x^k\) for \(k=1, 2, \dots, 10\). There are 10 such terms (\(x^1\) through \(x^{10}\)).
• Terms involving powers of y: These terms come only from the expansion of \((1+2y)^{10}\). They are of the form \(\binom{10}{j}(2y)^j\) for \(j=1, 2, \dots, 10\). There are 10 such terms (\(y^1\) through \(y^{10}\)).

Since terms with \(x\) and terms with \(y\) are linearly independent (cannot be combined), the total number of terms in the sum is the sum of the number of constant terms, the number of terms involving \(x\), and the number of terms involving \(y\).

Total number of terms = (Number of constant terms) + (Number of terms with \(x^k, k>0\)) + (Number of terms with \(y^j, j>0\))

Total number of terms = 1 + 10 + 10 = 21.

Therefore, there are 21 terms in the expansion of \((1 + 2x + x^2)^5 + (1 + 4y + 4y^2)^5\).

Summary of Term Counting

Conclusion on Number of Terms

By simplifying the initial expression and analyzing the terms generated by the binomial expansions of \((1+x)^{10}\) and \((1+2y)^{10}\), we determined that the terms involving \(x\) and terms involving \(y\) are distinct. Only the constant terms combine. This results in a total of 21 terms in the final expanded form.

Revision Table: Key Concepts

Additional Information: Generalizing Number of Terms

When adding two polynomials \(P_1\) and \(P_2\) that involve entirely different sets of variables (except possibly a constant term), the number of terms in \(P_1 + P_2\) is typically the sum of the number of terms in \(P_1\) and the number of terms in \(P_2\), minus one if both polynomials have a non-zero constant term (because the two constant terms combine into a single constant term). If only one has a constant term, or neither does, then the number of terms is simply the sum of the number of terms in each.

In our case, both \((1+x)^{10}\) and \((1+2y)^{10}\) have a constant term (\(\binom{10}{0} = 1\)). So, the total number of terms is (Terms in \((1+x)^{10}\)) + (Terms in \((1+2y)^{10}\)) - 1 (for the combined constant term).

Number of terms = 11 + 11 - 1 = 22 - 1 = 21.

This confirms our step-by-step calculation. The terms with \(x\) are distinct from the terms with \(y\). The only common 'type' of term is the constant term.