To address the given problem, we start by understanding the expansion of [tex]\((x + y)^4\)[/tex]:
[tex]\[
(x + y)^4 = K_1 x^4 + K_2 x^3 y + K_3 x^2 y^2 + K_4 xy^3 + K_5 y^4
\][/tex]
From the binomial theorem, we know the coefficients are the binomial coefficients:
[tex]\[
(x + y)^4 = \sum_{i=0}^{4} \binom{4}{i} x^{4-i} y^i
\][/tex]
Therefore,
[tex]\[
K_1 = \binom{4}{0} = 1, \quad K_2 = \binom{4}{1} = 4, \quad K_3 = \binom{4}{2} = 6, \quad K_4 = \binom{4}{3} = 4, \quad K_5 = \binom{4}{4} = 1
\][/tex]
Now, let's define the sums required for the question:
1. [tex]\(\mu \sum_{i=1}^5 K_i\)[/tex]
2. [tex]\(2 \sum_{i=2}^4 K_i\)[/tex]
3. [tex]\(3 \sum_{i=1}^4 K_i - 2 \sum_{i=3}^5 K_i\)[/tex]
### Step-by-step solution:
1. Given coefficients:
[tex]\[
K_1 = 1, \quad K_2 = 4, \quad K_3 = 6, \quad K_4 = 4, \quad K_5 = 1
\][/tex]
2. Calculating each required expression:
i. [tex]\(\mu \sum_{i=1}^5 K_i\)[/tex]
[tex]\[
\sum_{i=1}^5 K_i = K_1 + K_2 + K_3 + K_4 + K_5 = 1 + 4 + 6 + 4 + 1 = 16
\][/tex]
Assuming [tex]\(\mu = 1\)[/tex]
[tex]\[
\mu \sum_{i=1}^5 K_i = 1 \cdot 16 = 16
\][/tex]
ii. [tex]\(2 \sum_{i=2}^4 K_i\)[/tex]
[tex]\[
\sum_{i=2}^4 K_i = K_2 + K_3 + K_4 = 4 + 6 + 4 = 14
\][/tex]
[tex]\[
2 \sum_{i=2}^4 K_i = 2 \cdot 14 = 28
\][/tex]
iii. [tex]\(3 \sum_{i=1}^4 K_i - 2 \sum_{i=3}^5 K_i\)[/tex]
[tex]\[
\sum_{i=1}^4 K_i = K_1 + K_2 + K_3 + K_4 = 1 + 4 + 6 + 4 = 15
\][/tex]
[tex]\[
\sum_{i=3}^5 K_i = K_3 + K_4 + K_5 = 6 + 4 + 1 = 11
\][/tex]
[tex]\[
3 \sum_{i=1}^4 K_i - 2 \sum_{i=3}^5 K_i = 3 \cdot 15 - 2 \cdot 11 = 45 - 22 = 23
\][/tex]
### Final Results:
1. [tex]\(\mu \sum_{i=1}^5 K_i = 16\)[/tex]
2. [tex]\(2 \sum_{i=2}^4 K_i = 28\)[/tex]
3. [tex]\(3 \sum_{i=1}^4 K_i - 2 \sum_{i=3}^5 K_i = 23\)[/tex]
Thus, the values are:
(i) [tex]\( \mu \sum_{i=1}^5 K_i = 16 \)[/tex]
(ii) [tex]\( 2 \sum_{i=2}^4 K_i = 28 \)[/tex]
(iii) [tex]\( 3 \sum_{i=1}^4 K_i - 2 \sum_{i=3}^5 K_i = 23 \)[/tex]
