Repeat the procedure of ((b) starting from the identity (k+1)4-k4 = 4k +6k+4k+1 to derive a...

 

  

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Repeat the procedure of ((b) starting from the identity (k+1)4-k4 = 4k +6k+4k+1 to derive a formula for 13. (b) You probably also know the formula 1 k = n(n + 1)(2n+1). You can derive this formula as follows. Start by using the binomial theorem to prove the formula (k+1)3k3=3k + 3k+1. Writing this formula for k = 1,..., n you obtain: 23 13 3 12+3.1+1 33 23 3.22 +3.2+1 43 33 3.32 +3.3+1 (n+1)3 n 3 n+3. n+1 Now add these together to get: ... (n+1)313[1 + + n] + 3[1+...+n]+n. The left hand side is n +3n + 3n. The right hand side is 32 (which is 3 times the thing we are trying to compute) plus n(n+1)+n. So if you rearrange the formula, you can derive an expression for 12. Do this and simplify to see that you do indeed get n(n + 1)(2n+1). k=1 Repeat the procedure of ((b) starting from the identity (k+1)4-k4 = 4k +6k+4k+1 to derive a formula for 13. (b) You probably also know the formula 1 k = n(n + 1)(2n+1). You can derive this formula as follows. Start by using the binomial theorem to prove the formula (k+1)3k3=3k + 3k+1. Writing this formula for k = 1,..., n you obtain: 23 13 3 12+3.1+1 33 23 3.22 +3.2+1 43 33 3.32 +3.3+1 (n+1)3 n 3 n+3. n+1 Now add these together to get: ... (n+1)313[1 + + n] + 3[1+...+n]+n. The left hand side is n +3n + 3n. The right hand side is 32 (which is 3 times the thing we are trying to compute) plus n(n+1)+n. So if you rearrange the formula, you can derive an expression for 12. Do this and simplify to see that you do indeed get n(n + 1)(2n+1). k=1