(A): $$1^2 + 2^2 + 3^2 + ...... +49^2$$ = 40425

(R): Sum of the squares of the first n natural numbers is $$\frac{n(n + 1)(n + 2)}{6}$$

Solution

We know that sum of squares of first n naturals numbers is $$n(n+1)(2n+1)/6$$.

Condition A:

$$n=49.$$

so,$$1^2+2^2+3^2+......+49^2$$

$$=49×(49+1)(49×2+1)/6$$

$$=49×50×99/6$$

$$=40425.$$

Condition R:

$$n(n+1)(n+2)/6$$

$$=49×50×51/6$$

$$=20825.$$

which doesn't satisfy the given equation.

So,Option C is correct choice.

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