Question

1. Let $n \in \mathbb{N}$, $k \in \mathbb{N}$ such that $0 \le p \le n$ a. Find the term of $y^3$ in $(3x + 2y)^6$ b. Compute the sum of the combination terms $\sum_{k=0}^6 \binom{6}{k}$ c. Compute $\sum_{k=0}^n \binom{n}{k} + \sum_{k=0}^n (-1)^k \binom{n}{k}$ d. Show that $\binom{n}{k} = \binom{n-1}{k} + \binom{n-1}{k-1}$

          1.
Let $n \in \mathbb{N}$, $k \in \mathbb{N}$ such that $0 \le p \le n$
a. Find the term of $y^3$ in $(3x + 2y)^6$
b. Compute the sum of the combination terms $\sum_{k=0}^6 \binom{6}{k}$
c. Compute $\sum_{k=0}^n \binom{n}{k} + \sum_{k=0}^n (-1)^k \binom{n}{k}$
d. Show that $\binom{n}{k} = \binom{n-1}{k} + \binom{n-1}{k-1}$

1.
Let n ∈ℕ, k ∈ℕ such that 0 ≤ p ≤ n
a. Find the term of y^3 in (3x + 2y)^6
b. Compute the sum of the combination terms ∑k=0^6 6k
c. Compute ∑k=0^n nk + ∑k=0^n (-1)^k nk
d. Show that nk = n-1k + n-1k-1

Added by Jose Francisco H.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

Instant Answer

Solved by Expert Sri K

01/29/2023

Step 1

The term containing y^(3) will be when we choose y from 2y three times and x from 3x three times. This will happen when k = 3 in the binomial expansion formula. Therefore, the term will be: 6C3 * (3x)^(3) * (2y)^(3) = 20 * 27x^(3) * 8y^(3) = 4320x^(3)y^(3) Show more…

Show all steps

Thanks for your feedback!

Let ninN,kinN such that 0<=p<=n a. Find the term of y^(3) in (3x+2y)^(6) b. Compute the sum of the combination terms sum_(k=0)^6 ([6],[k]) c. Compute sum_(k=0)^n ([n],[k])+sum_(k=0)^n (-1)^(n)([n],[k]) d. Show that ([n],[k])=([n-1],[k])+([n-1],[k-1]) 1. Let n E N,k E N such that 0 < p < n a. Find the term of y3 in (3x + 2y) c. Compute EK=o(k) +=-1(K = += d. Show that