The square of the natural numbers are grouped as $$1^2,(2^2,3^2), (4^2,5^2,6^2),...,(..., 210^2)$$​

Which one of the following is correct in respect of the Question and the Statements?

Statement 1: The sum of the terms in the last group is 804660.

Statement 2: There are 20 groups in all.

Statement 3: The sum of all the terms till the last but one group is 2304415.

Correct option is D

Given: $$1^2$$, ($$2^2, 3^2$$), ( $$4^2, 5^2, 6^2$$), ........ (210²)

Formula Used:

The total number of terms from 1st to nth group = $$1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2}$$​​

Formula Used:

​$$\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$$

Where, n = Number of terms

Solution:

These are squares of the first 210 natural numbers => total terms = 210

$$\Rightarrow$$​ n(n+1) = 420

$$\Rightarrow$$​ n = 20

Statement I: The sum of the terms in the last group is 804660.

sum of the last group 20 terms, from 1912191^2191² to  2102210^2210² :

Last Term =  (191² + 192² + .........+ 210²)

=> (1² + 2² +..... + 210²) - (1² + 2² + .....+ 190²)​

=  $$\frac{210 \times 211 \times 421}{6} - \frac{190 \times 191 \times 381}{6}$$​​

=  $$\frac{18654510} {6}$$ -  $$\frac{13826490}{6}$$

= 3109085 -  2304415

= 804670

Therefore, 804670 $$\ne$$ 804660

Thus, this statement is false.

​Statement 2: There are 20 groups in all.

$$\frac{n(n+1)}{2} = 210 \Rightarrow n = 20 \text{ groups}$$

So, statement II is true.

Statement 3: The sum of all the terms till the last but one group is 2304415.​

Last Term =  (191² + 192² + .........+ 210²)

Sum of all the term except last term  =  1² + 2² + .....+ 190²​

=> $$\sum_{k=1}^{190} k^2 = \frac{190 \times 191 \times381}{6} = 2304415$$

So, the statement II is also correct.

Thus, the correct option is (d) Statements 2 and 3 are both correct, but Statement 1 is incorrect.