We can solve this problem using statement of principle of mathematical induction.
Statement of the Principle of Mathematical Induction:
Let P (n) be a statement involving the natural number ‘n’
such that
(i) P (1) is true i.e. P(n) is true for n =1 and
(ii) P (m+1) is true, whenever P(m) is true
i.e., P(m) is true ⇒ P(m+1) is true.
Then, P (n) is true for all natural numbers n.
By using these steps we can find the answer.
Given problem is Summation of series type
When n = 1, we have
++ =
++ =
=
= 1 ∊ N
We shall show that
(++) is a natural number for all n ∊ N.
Let P (n) be the statement :
++ is a natural number
Basic step :
For n = 1
++ =
++ =
=
= 1 ∊ N
Thus P (1) is true.
Induction step :
Assume that P 9k) is true that is
++ = m ... (i)
where m ∊ N
now, to prove P (k + 1) is true
(++) =
=
(++) +
(k4+2k3+2k2+k+) +
(k2+k+)+ = m +
(Using (i))
=
=
m+(k4+2k3+3k2+2k) + 1 ∊ N
[since m ∊ N and
(k4+2k3+3k2+2k) + 1 ∊ N]
Thus, P k + 1) is true whenever P (k) is true
Hence, by mathematical induction P (n) is true for all n ∊ N
i.e.,
(++) is a natural number for all n ∊ N.