Prove (A1 + A2 +...+ An)^T = A1^T + A2^T +...+ An^T using Mathimatical Induction.

clearly false, since

(a+b)^2 ≠ a^2+b^2

I suspect there's more to this besides the misspelling ...

I probably should have elaborated. This is a proof based on the Transposition of Matrices, and it requires the use of the theorem "(A + B)^T = A^T + B^T". It is not a series of integers being raised to a power.

Also, yes. I accidentally misspelled Mathematics.

well, if you have (A + B)^T = A^T + B^T then it is a cinch.

A1^T = A1^T (true for n=1
(A1+A2)^T = A1^T+A2^T (by the theorem)
matrix addition is associative, so
(A1+A2+A3)^T = ((A1+A2)+A3)^T
= (A1+A2)^T + A3^T
= A1^T+A2^T+A3^T
You can see how this can go directly from n=k to n=k+1

To prove the given statement using mathematical induction, we need to establish two things:

1. Base Case: Show that the statement holds true for the smallest possible value of "n".
2. Inductive Step: Assume that the statement holds true for some value of "n" and then prove that it also holds true for "n+1".

Let's go through both steps to prove the given statement:

1. Base Case (n = 1):
When n = 1, we have (A1)^T = A1^T, which is true.

2. Inductive Step:
Assume that the statement holds true for some arbitrary value of "n". That is,
(A1 + A2 + ... + An)^T = A1^T + A2^T + ... + An^T. ------ (1)

We need to prove that the statement holds true for "n + 1", which is:
(A1 + A2 + ... + An + An+1)^T = A1^T + A2^T + ... + An^T + (An+1)^T. ------ (2)

Starting with the left side of equation (2):
(A1 + A2 + ... + An + An+1)^T

By using the transpose property, we can rewrite this as:
(A1 + A2 + ... + An)^T + (An+1)^T

From our assumption in equation (1), we know that:
(A1 + A2 + ... + An)^T = A1^T + A2^T + ... + An^T

Substituting this into equation (2):
(A1^T + A2^T + ... + An^T) + (An+1)^T

Rearranging the terms using the associative property of addition:
A1^T + A2^T + ... + An^T + (An+1)^T

Therefore, we have derived equation (2) from our assumption, which proves the inductive step.

Based on the principle of mathematical induction, we can conclude that the given statement is true for all values of "n".