Moderate -0.3 This is a straightforward proof by induction with a standard divisibility result. The base case is trivial (5+22=27), and the inductive step requires routine algebraic manipulation to factor out 3 from 5^(k+1) + 2×11^(k+1) using the inductive hypothesis. While it's a Further Maths topic, it follows the standard template for divisibility proofs without requiring any novel insight or particularly complex algebra.
Question 9:
9 | When n = 1, 51 + 2111 = 27 div by 3
Assume u = 5k + 211k is div by 3
k
u = 5k+1 + 211k+1
k+1 | B1*
M1
M1 | 2.1
2.1 | or 5k + 211k = 3m
= 5(u 211k) + 2211k
k
= 5u + 1211k
k | M1
A1 | 1.1b | substituting for 5k
or 15m + 1211k
or u = 5k+1 + 11(u 5k)
k+1 k | M1 | substituting for 11k | 5.5k + 11(3m 5k)
= 11u 65k
k | A1 | or 33m 65k
or u + u = 5k+1 + 211k+1 + 5k + 211k
k+1 k
= 6 5k + 24 11k | M1
A1 | 1.1b | adding u to u
k k+1
As u div by 3, u div by 3
k k+1
So if true for n = k, true for n = k+1. As true
for n = 1, true for all positive integers n | A1*
A1dep
[7] | 2.2a
2.4 | dep * marks
9 Prove by induction that $5 ^ { n } + 2 \times 11 ^ { n }$ is divisible by 3 for all positive integers $n$.
\hfill \mbox{\textit{OCR MEI Further Pure Core 2019 Q9 [7]}}