AP Maths
Induction
Steps:
1. Prove true for n=1
2. Assume true for n=k [where k is any arbitrary value]
3. Prove true for n=k+1
4. Conclusion - if it is true for n=1 and n=1+1=2, then it is now true for n=2 and n=2+1=3...
therefore true when n∈ℝ
*must include conclusion
- If you prove something is true for one value, and it is also true for that value + 1, then it
is true for all numbers
Examples:
a. Prove that:
n
n
∑ ❑i = 2 [n+1]
i=1
n
1+2+3….n = [n+1]
2
1. Prove true for n=1
LHS = 1
RHS = ½ [1+1]
∴ true for n=1
2. Assume true for an arbitrary value n=k
1+2+3...+k= k/2[k+1]
3. Prove true for n=k+1
[k +1]
RTP: [require to prove] RHS = [k+1+1]
2
[k +1][k + 2]
=
2
*take RHS from step 2 and add on one term → [n=k + i=k+1]
RHS = k/2[k+1] + k+1
2❑
[k +k +2 k +2]
=
2
[ k +1][k + 2] equal to step 3
= →
2
4. It is true for n=1 and n=k+1
∴ It is true for n=2 and n=k+1 ∴ it is true for n=3 and so on
n
n
∴ ∑ ❑i = 2 [n+1] for n∈ℝ
i=1
1
, AP Maths
b. Prove that:
n
n
∑ ❑[3i+2] = 2 [3n+7]
i=1
n
5+8+11...n = [3n+7]
2
1. Prove true for n=1
LHS = 5
RHS = ½ [3(1)+1]
=5
∴ true for n=1
2. Assume true for an arbitrary value n=k
5+8+11…+[3k+2] → sub in place of i = k/2[3k+7]
3. Prove true for n=k+1
[k +1]
RTP: [require to prove] RHS = [3(k+1)+7]
2
3 k 2 +13 k +10
=
2
*take RHS from step 2 and add on one term → [n=k + i=k+1] [i = after …]
RHS = k/2[3k+7] + [3(k+1)+2]
2
3 k +13 k +10 equal to step 3
= →
2
4. It is true for n=1 and n=k+1
∴ It is true for n=2 and n=k+1 ∴ it is true for n=3 and so on
n
n
∴ ∑ ❑[3i+2] = 2 [3n+7] for n∈ℝ
i=1
c. Prove that:
2 2 2 n[ 4 n2−1]
1 +3 +5 …[2n-1]2 =
3
2
(1)[4 (1) −1]
1. LHS = 1 RHS =
3
=1
2
2 2 2 k [4 k −1]
2
2. Assume 1 +3 +5 …[2k-1] =
3
k +1[4 (k +1)2−1]
3. RTP:
3
2
Induction
Steps:
1. Prove true for n=1
2. Assume true for n=k [where k is any arbitrary value]
3. Prove true for n=k+1
4. Conclusion - if it is true for n=1 and n=1+1=2, then it is now true for n=2 and n=2+1=3...
therefore true when n∈ℝ
*must include conclusion
- If you prove something is true for one value, and it is also true for that value + 1, then it
is true for all numbers
Examples:
a. Prove that:
n
n
∑ ❑i = 2 [n+1]
i=1
n
1+2+3….n = [n+1]
2
1. Prove true for n=1
LHS = 1
RHS = ½ [1+1]
∴ true for n=1
2. Assume true for an arbitrary value n=k
1+2+3...+k= k/2[k+1]
3. Prove true for n=k+1
[k +1]
RTP: [require to prove] RHS = [k+1+1]
2
[k +1][k + 2]
=
2
*take RHS from step 2 and add on one term → [n=k + i=k+1]
RHS = k/2[k+1] + k+1
2❑
[k +k +2 k +2]
=
2
[ k +1][k + 2] equal to step 3
= →
2
4. It is true for n=1 and n=k+1
∴ It is true for n=2 and n=k+1 ∴ it is true for n=3 and so on
n
n
∴ ∑ ❑i = 2 [n+1] for n∈ℝ
i=1
1
, AP Maths
b. Prove that:
n
n
∑ ❑[3i+2] = 2 [3n+7]
i=1
n
5+8+11...n = [3n+7]
2
1. Prove true for n=1
LHS = 5
RHS = ½ [3(1)+1]
=5
∴ true for n=1
2. Assume true for an arbitrary value n=k
5+8+11…+[3k+2] → sub in place of i = k/2[3k+7]
3. Prove true for n=k+1
[k +1]
RTP: [require to prove] RHS = [3(k+1)+7]
2
3 k 2 +13 k +10
=
2
*take RHS from step 2 and add on one term → [n=k + i=k+1] [i = after …]
RHS = k/2[3k+7] + [3(k+1)+2]
2
3 k +13 k +10 equal to step 3
= →
2
4. It is true for n=1 and n=k+1
∴ It is true for n=2 and n=k+1 ∴ it is true for n=3 and so on
n
n
∴ ∑ ❑[3i+2] = 2 [3n+7] for n∈ℝ
i=1
c. Prove that:
2 2 2 n[ 4 n2−1]
1 +3 +5 …[2n-1]2 =
3
2
(1)[4 (1) −1]
1. LHS = 1 RHS =
3
=1
2
2 2 2 k [4 k −1]
2
2. Assume 1 +3 +5 …[2k-1] =
3
k +1[4 (k +1)2−1]
3. RTP:
3
2
