3. Prove by induction that∑_(r=1)^n▒〖r(r+4)=1/6 n(n+1)(2n+13)〗.
5. It is given that u_1=1 and u_(n+1)=3u_n+2n-2 where n is a positive integer.
Prove, by induction, that u_n=3^n/2-n+1/2.
14. The rth term of a sequence is given by u_r=r(3r+1),r=1,2,3,…
i) Write down the values of ∑_(r=1)^n▒u_r for n=1,2,3, and 4.
ii) Make a conjecture for a formula for ∑_(r=1)^n▒u_r , giving your answer in the form nf(n), where f(n) is a function of n.
iii) Prove by induction a formula for ∑_(r=1)^n▒u_r .
If the text isn't displaying properly, please use my screenshot.
i44.tinypicDOTcom/66dwnb.png
Sorry about the inconvenience, and thanks!
3. Prove by induction that ∑(r=1)^(n) [ r(r+4) ] = 1/6 n(n+1)(2n+13).
Step 1: Base case (n = 1)
- We need to prove that ∑(r=1)^(1) [ r(r+4) ] = 1/6 (1)(2)(15).
- For n = 1, the left-hand side becomes: ∑(r=1)^(1) [ r(r+4) ] = 1(1+4) = 5.
- The right-hand side becomes: 1/6 (1)(2)(15) = 15/6 = 5/2.
- So, the base case holds true.
Step 2: Inductive hypothesis
- Assume that the statement is true for some positive integer value k, i.e.,
∑(r=1)^(k) [ r(r+4) ] = 1/6 k(k+1)(2k+13).
Step 3: Inductive step
- We need to prove that the statement holds for k+1, i.e.,
∑(r=1)^(k+1) [ r(r+4) ] = 1/6 (k+1)(k+2)[2(k+1)+13].
- We have: ∑(r=1)^(k+1) [ r(r+4) ] = ∑(r=1)^(k) [ r(r+4) ] + (k+1)((k+1)+4)
(by expanding one more term in the summation).
- Using the inductive hypothesis, we can rewrite the above equation as:
1/6 k(k+1)(2k+13) + (k+1)((k+1)+4)
= (k+1)(1/6 k(2k+13) + ((k+1)+4))
= (k+1)(1/6)(2k^2 + 13k + 6k + 6)
= (k+1)(1/6)(2k^2 + 19k + 6)
= (k+1)(1/6)(2k^2 + 13k + 6k + 6)
= (k+1)(1/6)(2k(k+1) + 13(k+1) + 6)
= 1/6 (k+1)(k+2)(2k + 13 + 6).
- Simplifying the above expression, we get: 1/6 (k+1)(k+2)(2k + 19).
- This is equivalent to: 1/6 (k+1)(k+2)(2(k+1) + 13).
- Therefore, the statement holds for k+1.
Step 4: Conclusion
- By the principle of mathematical induction, we can conclude that
∑(r=1)^(n) [ r(r+4) ] = 1/6 n(n+1)(2n+13) for all positive integers n.