Question

the binomial theorem states that for any real numbers a and b (a b)n=∑nk=0(nk)an−kbk, for any integer n ≥0 use this theorem to show that for any integer n ≥0m ∑nk=0(−1)k(nk)3n−k2k=1

Answer 1

Explanation:

We can use the Binomial Theorem to show this by letting a=3 and b=2 in the formula:

(a+b)^n = ∑(n choose k) a^(n-k) b^k

Substituting the values of a and b, we get:

(3+2)^n = ∑(n choose k) 3^(n-k) 2^k

5^n = ∑(n choose k) 3^(n-k) 2^k

Multiplying both sides by (-1)^n, we get:

(-1)^n 5^n = ∑(n choose k) (-1)^n 3^(n-k) 2^k

(-1)^n 5^n = ∑(n choose k) (-1)^k 3^(n-k) 2^k

Using the property that (n choose k) = (n choose n-k), we can simplify the expression:

(-1)^n 5^n = ∑(n choose n-k) (-1)^(n-k) 3^(k) 2^(n-k)

(-1)^n 5^n = ∑(n choose k) (-1)^(n-k) 3^(k) 2^(n-k)

We recognize the sum on the right-hand side as the expansion of (3-2)^n:

(-1)^n 5^n = (3-2)^n = ∑(n choose k) (-1)^(n-k) 3^(k) 2^(n-k)

Rearranging, we get:

∑(n choose k) (-1)^k 3^(n-k) 2^k = 5^n

Dividing both sides by 5^n, we get:

∑(n choose k) (-1)^k (3/5)^(n-k) (2/5)^k = 1

We recognize the left-hand side as a binomial expansion with coefficients (n choose k) and terms (3/5)^(n-k) and (2/5)^k. Therefore, the sum of these terms must equal 1, by the Binomial Theorem. This verifies the result.