Farhanthestudent005
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- 2022
Need help with these two...thanks
Questions help needed (1 Viewer)
- Thread starter Farhanthestudent005
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Need help with these two...thanks
Give the questions a go first. So we can see what you have tried
For this question
-2)=\frac{(k+1)(3(k+1)-1)}{2})
subbing in n=k+1
+(3(k+1)-2)\\\\LHS=\frac{k(3k-1)}{2}+(3(k+1)-2))
Using the inductive statement that would have been stated if it was worked out properly.
+6k+2}{2}\\\\LHS=\frac{3k^2+5k+2}{2}\\\\LHS=\frac{3k^2+3k+2k+2}{2}\\\\LHS=\frac{(k+1)(3k+2)}{2}\\\\LHS=\frac{(k+1)(3(k+1)-1)}{2}\\\\LHS=RHS)
Therefore by mathemtatical induction the following statement blah blah blahg....
Lets start off with part 3 of maths induction, let n=k+1
}{1-r})
}{1-r}+T_1r^k\\\\LHS=\frac{T_1(1-r^{k})+(1-r)T_1r^k}{1-r}\\\\LHS=\frac{T_1(1-r^k+(1-r)r^k)}{1-r}\\\\LHS=\frac{T_1(1-r^{k+1})}{1-r}\\\\LHS=RHS)
Add the final concluding sentence...
Skipping a few steps we need to prove
Therefore by mathemtatical induction the following statement blah blah blahg....
Level 2 is a bit harder
Lets start off with part 3 of maths induction, let n=k+1
Add the final concluding sentence...