We need to find the numerically greatest term in the expansion of \( (5 + 3x)^6 \). The general term in the binomial expansion of \( (5 + 3x)^6 \) is: \( T_r = \binom{6}{r} 5^{6-r} (3x)^r \)
Step 1: Substitute \( x = 1 \) into the general term: \( T_r = \binom{6}{r} 5^{6-r} 3^r \)
Step 2: The term will be greatest when the powers of 3 and 5 are balanced. After solving, the greatest term occurs when \( r = 3 \), and the value is \( 3^3 \times 5^5 \).
Let $ (1 + x + x^2)^{10} = a_0 + a_1 x + a_2 x^2 + ... + a_{20} x^{20} $. If $ (a_1 + a_3 + a_5 + ... + a_{19}) - 11a_2 = 121k $, then k is equal to _______
The term independent of $ x $ in the expansion of $$ \left( \frac{x + 1}{x^{3/2} + 1 - \sqrt{x}} \cdot \frac{x + 1}{x - \sqrt{x}} \right)^{10} $$ for $ x>1 $ is: