Dayo from Queen Elizabeth Grammar School, Horncastle proved the result using modulus arithmetic. This is his solution but written so that you do not need to know about modulus arithmetic to understand the solution.

We must show that 3(3n+4)+7(2n+1) is a multiple of 11 for all positive integer values of n.

Let
f(n)
= 3(3n+4)+7(2n+1)
= 27n×81 + 49n ×7
= (22 +5)n(77+4)+(44+5)n(11-4).
As all the terms except the term 5n of the binomial expansions (22+5)n and (44+5)n involve multiples of 11, it follows that for some integer k:
f(n) = 11k + 5n ×4 + 5n × (-4) = 11k +5n(4-4) = 11k.
Hence, for any integer value of n, f(n) (as stated above) is divisible by eleven.

Good solutions using the axiom of mathematical induction were submitted by JiaMin from St Clare's School Oxford, Simba from Bury Grammar School, James from Aston Comprehensive School, Andrei from Tudor Vianu National College, Bucharest, Romania and Ruth from Manchester High School for Girls.

This is James's solution:

To prove P(n): 3(3n+4) + 7(2n+1) is a multiple of 11.

Consider P(1): 37 + 73 = 2530, which is a multiple of 11. Therefore P(1) is true.

Assume P(k) is true. Therefore 3(3k+4) + 7(2k+1) is a multiple of 11. Consider
P(k+1): 3(3(k+1)+4) + 7(2(k+1)+1)
= 3(3k+7) + 7(2k+3)
= 27(3(3k+4)) + 49(7(2k+1))
= 27(3(3k+4) + 7(2k+1)) + 22(7(2k+1))
.
As 3(3k+4) + 7(2k+1) is a multiple of 11 and 22 is a multiple of 11 therefore 3(3(k+1)+4) + 7(2(k+1)+1) is a multiple of 11, that is P(k+1) is true.

We have shown P(1) is true and P(k) implies P(k+1) therefore, by the axiom of mathematical induction, P(n) is true for all values of n.