Question

(complete solution), can someone help me here please, i need ur help asap thank you!

Formula
Step 1: n=1
Step 2: n=k
Step 3: n= k+ 1​

Answer 1

See proof below

Explanation:

We will denote the sum of n elements of the sequence as S(n)

What is proof by induction?

We first assume that for n=1 , S(1) is true
This step is called the basis

Then for n= k ≥ 1, prove S(n) is true and then prove S(k+1) also true. This is called the induction step

For this specific problem, we assume S(1) is true


`\mbox{\large S(1) = 2}`

`\mbox{\large S(k) = 2 + 4 + 6 + ..... + 2k = k(k+1) }`

We are using 2k because the values are twice the index values 1, 2, 3...
2.1, 2.1, 2.3 ...2k

`\mbox{\large We now show }`
`\mbox{\large S(k + 1) }`
`\mbox{\large is true }`

`\mbox {\large S(k + 1) = S(k) + 2(k + 1)}`

`\mbox{\large 2+4+6+...+2k+2(k+1)}\\\\\mbox{\large=k(k+1)+2(k+1)}\\\\\mbox{\large=(k+1)(k+2)}\\\\= \mbox{\large (k+1)(k+1+1)}}\\\\`

`\mbox{\large Substituting n for k + 1 we get }`

`\mbox{\large S(n) = n(n+1)}`

I hope the proof is comprehensible. If not please ask for clarifications

Answer 2

the proof is below

Explanation:

You want a proof by induction of the formula for the sum of consecutive even numbers.

Step 1: base case

The given formula is ...

Sn = n(n+1)

For n=1, the one-term sequence is the first term: 2. Its sum is 2.

2 = (n)(n+1) = (1)(1+1) = 2 . . . . . true

Step 2: n = k

For this step, we assume the formula is correct. Then ...

Sk = k(k+1)

Step 3: n = k+1

The sequence sum for k+1 terms is ...

2 + 4 + 6 + ... + 2k = k(k+1) . . . . . assumption from step 2

(2 + 4 + 6 + ... + 2k) +2(k+1) . . . . . one more term added; sum of k+1 terms

k(k+1) +2(k+1) . . . . . . . . substitute assumed value for (2 + 4 + ...)

(k +2)(k +1) . . . . . . . . . . factor out k+1

And the formula for k+1 terms is ...

= (k+1)((k+1) +1) . . . . . substitute (k+1) for n

= (k+1)(k+2) . . . . . . . perform the addition

Note that the sum of terms is (k+2)(k+1), and is identical to the formula for the same sum: (k+1)(k+2). (Remember, multiplication is commutative.)

(k+2)(k+1) = (k+1)(k+2) . . . . . QED