Let `A_(n) = a_(1) + a_(2) + "……" + a_(n), B_(n) = b_(1) + b_(2) + b_(3) + "…." + b_(n), D_(n) = c_(1) + c_(2) + "….." + c_(n)` and `c_(n) = a_(1)b_(n) + a_(2)b_(n-1) + "……." + a_(n)b_(1)Aan in N`. Using mathematical induction , prove that
(a) `D_(n) = a_(1)B_(n) + a_(2)B_(n-1) + "....."+a_(n)B_(1) = b_(1)A_(n) + b_(2)A_(n-1) + "......"+b_(n)A_(1) AA n in N`
(b) `D_(1) + D_(2) + "......"+ D_(n) = A_(1)B_(n) + A_(2)B_(n-1) + "....." + A_(n)B_(1) AA n in N`

Explanation

Let ` P(n) : D_(n) a_(n)B_(n) + a_(n)B_(n-1)+"……."+a_(n)B_(1)`
For `n = 1, P(1) = D_(1) = a_(1)B_(1) = a_(2)b_(1) = c_(1)`
Thus, ` P(1)` is true.
Let `P(n)` be true for `n = k`
i.e., `P(k) = D_(k) = a_(1)B_(k)= a_(1)B_(k) + a_(2)B_(k-1) + "....."+ a_(k)B_(1)`
Now, `D_(k+1) = D_(k) + c_(k+1)`
`= a_(1)B_(k) + a_(2)B_(k-1) + "......"+a_(k)B_(1) +a_(1)b_(k+1) + a_(2)b_(k) + "....."+a_(k+1)b_(1)`
`= a_(1)(B_(k)+b_(k+1))+ a_(2)(B_(k-1)+b_(k)) + "......" + a_(k)(B_(1) + b_(2)) + a_(k+1)b_(1)` ltbrlt `= a_(1)B_(k+1)+a_(2)B_(2)+"......."+a_(k)B_(2) +a_(k+1)B_(1)`.
Thus, `P(k+1)` is also true whenever `P(k)` is true.
So, by the principle of mathematical induction `P(n)` is true
for any natural number n.
Let `P_(1)(n) : D_(n) = b_(1)A_(n) + b_(2)A_(n-1)+"......."+b_(n)A_(1)`
For `n = 1, P_(1)(1) = D_(1) = b_(1)A_(1) = b_(1)a_(1) = c_(1)`
Thus, `P_(1)(1)` is true.
Let, `P_(1)(n)` be true for `n = k`.
i.e, `P_(1)(k) = b_(1)A_(k) + b_(2)A_(k-1) + "....." + b_(k)A_(1)`
Now, `D_(k+1) = D_(k) +c_(k+1)`
`= b_(1)A_(k) + b_(2)a_(k-1) + "......." + b_(k)A_(1) + a_(1)b_(k+1) + a_(2)b_(k) + "....." + a_(k+1)b_(1)`
`= b_(1)(A_(k) + a_(k+1))b_(2)(A_(k-1)+a_(k))+"......"+b_(k)(A_(1) + a_(2)) + b_(k+1)a_(1)`
`= b_(1)A_(k+1) + b_(2)A_(k) + "......"+b_(k)A_(2) + b_(k+1)A_(1)`
Thus, `P_(1)(k+1)` is also true.
So, by the principle of mathematical induction `P_(1)(n)` is true for any natural number n.
(b) Say, `P_(2)(n) = D_(1) + D_(2) + "......"+D_(n) = A_(1)B_(n) + A_(2)B_(n-1) + "......" + A_(n)B_(2)`
For `n = 1, P_(2)(1) = D_(1) = A_(1)B_(1) = a_(1)b_(1) = c_(1)`
Thus, `P_(2)(1)` is true .
Let `P_(2)(n)` be true for `n = k`
i.e, ` P_(2)(k) = A_(1)B_(k) + A_(2)B_(k+1) + "......" + A_(k)B_(1)`
Now, `P_(2)(k+1)`
`= P_(2)(k) + D_(k+1)`
`= A_(1)B_(k) + A_(2)B_(k-1) +"........"+)A_(k)B_(1) +b_(1)A_(k+1) + b_(2)A_(k) + "......" + b_(k)A_(2) + b_(k+1)A_(1)`
`= A_(1)(B_(k) + b_(k+1)) + A_(2)(B_(k-1)+b_(k)) + "....." + A_(k)(B_(1) + b_(2)) + A_(k+1)b_(1)`
` = A_(1)B_(k+1) + A_(2)B_(k) + "....." + A_(k)B_(2) + A_(k+1)B_(1)`
Thus, `P_(2)(k+1)` is also true.
So, by the principle of mathematical induction, `P_(2)(n)` is true for any natural number n.