Standard +0.3 This is a straightforward proof by induction with a given formula to prove. The base case is trivial (substitute n=1), and the inductive step requires only algebraic manipulation of the recurrence relation with no conceptual challenges. Slightly easier than average since the formula is provided and the algebra is routine.
5 You are given that \(u _ { 1 } = 5\) and \(u _ { n + 1 } = u _ { n } + 2 n + 4\).
Prove by induction that \(u _ { n } = n ^ { 2 } + 3 n + 1\) for all positive integers \(n\).
Question 5:
5 | When n = 1, n2 + 3n + 1 = 12 + 3×1 + 1 = 5
Assume true for n = k u = k2 + 3k + 1
k
Then u = u + 2k + 4
k+1 k
= k2 + 3k + 1 + 2k + 4
= (k + 1)2 + 3k + 4
= (k + 1)2 + 3(k + 1) + 1
So if true for n = k, then true for n = k + 1
As true for n = 1, true for all n. | B1
B1
M1
A1*
B1*dep
B1cao
[6] | 1.1
2.1
1.1
2.1
2.2a
2.4 | dep A1*
5 You are given that $u _ { 1 } = 5$ and $u _ { n + 1 } = u _ { n } + 2 n + 4$.\\
Prove by induction that $u _ { n } = n ^ { 2 } + 3 n + 1$ for all positive integers $n$.
\hfill \mbox{\textit{OCR MEI Further Pure Core AS 2020 Q5 [6]}}