Statement-1: `1^(2)+2^(2)+....+n^(2)=(n(n+1)(2n+1))/(6)"for all "n in N`
Statement-2: `1+2+3....+n=(n(n+1))/(2),"for all"n in N`

Options

A. statement 1 is true, statement 2 is false.
B. statement 1 is false, statement 2 is true.
C. statement 1 is true, statement 2 is true but statement 2 is not correct explaination of statement 1.
D. statement 1 is true, statement 2 is true and statement 2 is correct explaination of statement 1.

Answer

To solve the problem, we will use the Principle of Mathematical Induction to prove both statements.

### Step 1: Proving Statement 1
**Statement 1:** \( 1^2 + 2^2 + \ldots + n^2 = \frac{n(n+1)(2n+1)}{6} \) for all \( n \in \mathbb{N} \).

**Base Case:** For \( n = 1 \):
\[ 1^2 = 1 \]
The right side becomes:
\[ \frac{1(1+1)(2 \cdot 1 + 1)}{6} = \frac{1 \cdot 2 \cdot 3}{6} = 1 \]
Both sides are equal, so the base case holds.

**Inductive Step:** Assume true for \( n = k \):
\[ 1^2 + 2^2 + \ldots + k^2 = \frac{k(k+1)(2k+1)}{6} \]
Now we need to prove it for \( n = k + 1 \):
\[ 1^2 + 2^2 + \ldots + k^2 + (k+1)^2 = \frac{k(k+1)(2k+1)}{6} + (k+1)^2 \]
Combine the terms:
\[ = \frac{k(k+1)(2k+1)}{6} + \frac{6(k+1)^2}{6} = \frac{k(k+1)(2k+1) + 6(k+1)^2}{6} \]
Factor out \( (k+1) \):
\[ = \frac{(k+1)(k(2k+1) + 6(k+1))}{6} \]
Simplifying the expression inside the parentheses:
\[ = \frac{(k+1)(2k^2 + k + 6k + 6)}{6} = \frac{(k+1)(2k^2 + 7k + 6)}{6} \]
Factoring \( 2k^2 + 7k + 6 \):
\[ = \frac{(k+1)(k+2)(2k+3)}{6} \]
Thus, we have shown:
\[ 1^2 + 2^2 + \ldots + (k+1)^2 = \frac{(k+1)((k+1)+1)(2(k+1)+1)}{6} \]
This completes the inductive step, proving Statement 1.

### Step 2: Proving Statement 2
**Statement 2:** \( 1 + 2 + 3 + \ldots + n = \frac{n(n+1)}{2} \) for all \( n \in \mathbb{N} \).

**Base Case:** For \( n = 1 \):
\[ 1 = \frac{1(1+1)}{2} = 1 \]
Both sides are equal, so the base case holds.

**Inductive Step:** Assume true for \( n = k \):
\[ 1 + 2 + 3 + \ldots + k = \frac{k(k+1)}{2} \]
Now we need to prove it for \( n = k + 1 \):
\[ 1 + 2 + 3 + \ldots + k + (k+1) = \frac{k(k+1)}{2} + (k+1) \]
Combine the terms:
\[ = \frac{k(k+1) + 2(k+1)}{2} = \frac{(k+1)(k + 2)}{2} \]
Thus, we have shown:
\[ 1 + 2 + 3 + \ldots + (k+1) = \frac{(k+1)((k+1)+1)}{2} \]
This completes the inductive step, proving Statement 2.

### Conclusion
Both statements are proven true using the Principle of Mathematical Induction.