Question

The number of nonzero terms in the expansion of \[ (1 + 3\sqrt{2}x)^9 + (1 - 3\sqrt{2}x)^9 \] is:

• A. 2
• B. 3
• C. 4
• D. 5

Hint:

For expressions of the form \( (a+b)^n + (a-b)^n \), only even power terms remain nonzero due to the cancellation of odd power terms.

Answer 1

The Correct Option is D

We are asked to find the number of nonzero terms in the expansion of the given expression: \[ (1 + 3\sqrt{2}x)^9 + (1 - 3\sqrt{2}x)^9 \] We begin by expanding both binomials using the binomial theorem. The binomial expansion of \( (a + b)^n \) is given by: \[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \] Thus, the expansions of both binomials are: \[ (1 + 3\sqrt{2}x)^9 = \sum_{k=0}^{9} \binom{9}{k} 1^{9-k} (3\sqrt{2}x)^k = \sum_{k=0}^{9} \binom{9}{k} (3\sqrt{2})^k x^k \] and \[ (1 - 3\sqrt{2}x)^9 = \sum_{k=0}^{9} \binom{9}{k} 1^{9-k} (-3\sqrt{2}x)^k = \sum_{k=0}^{9} \binom{9}{k} (-3\sqrt{2})^k x^k \] Now, adding these two expansions together: \[ (1 + 3\sqrt{2}x)^9 + (1 - 3\sqrt{2}x)^9 \] We observe that terms with odd powers of \( x \) will cancel each other out, because the terms involving \( 3\sqrt{2}x \) and \( -3\sqrt{2}x \) will have opposite signs. Only the terms with even powers of \( x \) will remain nonzero. The terms with even powers of \( x \) are for \( k = 0, 2, 4, 6, 8 \). Therefore, there are 5 nonzero terms in the expansion. Thus, the number of nonzero terms is \( {5} \).