Question

Question 6: (a) (4 mark) Show that for n ? ?, n(1 + x)^{n-1} = ?_{k=1}^{n} k inom{n}{k} x^{k-1}. [Hint: you may find it useful to look at the expansion of (1 + x)^n with the Binomial theorem] (b) (1 mark) Prove that n × 2^{n-1} = inom{n}{1} + 2 inom{n}{2} + ? + n inom{n}{n} = ?_{k=1}^{n} k inom{n}{k} (c) (6 marks) Show that ?_{k=1}^{n} k^2 inom{n}{k} = n(n + 1) × 2^{n-2}.

          Question 6:
(a) (4 mark) Show that for n ? ?,
n(1 + x)^{n-1} = ?_{k=1}^{n} k inom{n}{k} x^{k-1}.
[Hint: you may find it useful to look at the expansion of (1 + x)^n with the Binomial theorem]
(b) (1 mark) Prove that
n × 2^{n-1} = inom{n}{1} + 2 inom{n}{2} + ? + n inom{n}{n} = ?_{k=1}^{n} k inom{n}{k}
(c) (6 marks) Show that
?_{k=1}^{n} k^2 inom{n}{k} = n(n + 1) × 2^{n-2}.

Question 6:
(a) (4 mark) Show that for n ? ?,
n(1 + x)^n-1 = ?k=1^n k inomnk x^k-1.
[Hint: you may find it useful to look at the expansion of (1 + x)^n with the Binomial theorem]
(b) (1 mark) Prove that
n × 2^n-1 = inomn1 + 2 inomn2 + ? + n inomnn = ?k=1^n k inomnk
(c) (6 marks) Show that
?k=1^n k^2 inomnk = n(n + 1) × 2^n-2.

Added by Lidia S.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Adi S

07/21/2023

Step 1

We need to show that this equation holds true for all $n \in \mathbb{N}$. Using the Binomial theorem, we can expand $(1+r)^{n-1}$ as follows: $(1+r)^{n-1} = \binom{n-1}{0} + \binom{n-1}{1}r + \binom{n-1}{2}r^2 + \cdots + \binom{n-1}{n-1}r^{n-1}$ Now, let's Show more…

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Question 6: (a) Show that for n ∈ ℕ, n(1 + x)^{n-1} = ∑_{k=1}^{n} k inom{n}{k} x^{k-1}. [Hint: you may find it useful to look at the expansion of (1 + x)^n with the Binomial theorem] (b) Prove that n × 2^{n-1} = inom{n}{1} + 2 inom{n}{2} + ⋯ + n inom{n}{n} = ∑_{k=1}^{n} k inom{n}{k} (c) Show that ∑_{k=1}^{n} k^2 inom{n}{k} = n(n + 1) × 2^{n-2}.