Question

Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false. 4-6+5-7+6-8 + ... + 4n( 4n + 2) = 4(4n+1)(8n+7)/6

Answer 1

Could you rewrite that expression? I'm confused on what the inductive formula is.

Answer 2

can you rewrite the sum of the terms pls

as it appears that it fails the test for n = 1

\[4(4 + 2) = \frac{4(4 + 1)(8 + 7)}{6}\]

Answer 3

4-6+5-7+6-8

Answer 4

I meant the expression to the farmost right of the equal sign. Is it what campbell_st posted?

Answer 5

yes

Answer 6

what is the sum term... is it correct...?

\[= \frac{4(4n + 1)(8n+7)}{6}\]

Answer 7

yea there it is he got it

Answer 8

ok... so the 1st thing to so is prove it correct for n = 1

so the 1st term is equal to the sum of 1 term

\[T_{1} = 4\times 1(4 \times 1 + 2) = 24\]

now look at the sum when n = 1

\[S_{1} = \frac{4(4 \times 1 + 1)\times (8 \times 1 + 7)}{6} = 50\]

so there would appear to be a mistake somewhere... if you are to prove it by induction

Answer 9

so it is false

Answer 10

well it would appear to fail the test where n = 1

Answer 11

Thanks

Answer 12

wait what about k ? @campbell_st

Answer 13

well you assume its true for n = k so \[S_{k} = \frac{4(4k + 1)(8k + 7)}{6}\]

but there is no use looking at n = k since it failed for n = 1