1.2 2,7,18 Prove 13 + 2³ + ... +N³ = [= N(N+1)]³² for all NEN Step 1 show N=1 Step 2 P=1 [HD] [²]²=1 Assume KEN 13³ +2³+ K³ = [ {K/K+1)] ²³ step 3 Apply K+1 15+2³+ +K³ + (K+1)³ = [K(K+1)] ² + (4+1) ³ InductioN (A+)K]" + (+1)(K+1)/K+1) (+)² (K+1)(h+1)K)/(k) + (K+1) / K+1)(K+1) = [ (h+1)] ²³ (1){(k) + (K+1) (K+1) {K+1) I Don't see how to go to ==+/(x+1)X(X+2)] ²

Using Wilson's Theorm, Fermat's Little Theorm and The Pollard Factorization Metod how could I solve Quetion 6 in Section 6.1

5:56 Telegram k + k+1 k+2 a) k+1 1 (k+1)((k+1)+1) k+1 k+2 k+1 b) k+3 d) k+1 k docs.google.com 46

3. In this problem, you may use the following formulae without further justification: 2 ^²³ − ( ² - - (n(n + ¹)) ². 2 £* k k=1 n(n+1) 2 12 k=1 k² (a) Find a formula in n for each of (b) Find a formula in n for each of Σ(2k) and Σ(2k - 1). k=1 k=1 n n(n+1)(2n +1) 6 (c) Find a formula in n for each of n Σ(2k)² and Σ(2k-1)². TL k=1 TL (d) Only use the formulae above, to compute Σ(2k)³ and Σ(2k-1)³. 1³22 +33 4²+ +99³-100², which is the sum of cubes of all odd integers no greater than 100, minus the sum of squares of all even integers no greater than 100.