$$Statement-1$$: The sum of the series $$1+(1+2+4)+(4+6+9)+(9+12+16)+$$…$$+(361+380+400)$$ is $$8000Statement-2$$: ∑$$k=1n$$ $$k3$$−$$(k$$−$$1)3=n3$$ for each natural number n.

Question

$$Statement-1$$: The sum of the series $$1+(1+2+4)+(4+6+9)+(9+12+16)+$$…$$+(361+380+400)$$ is  $$8000Statement-2$$: ∑$$k=1n$$ $$k3$$−$$(k$$−$$1)3=n3$$ for each natural number n.

Infinity Learn · Accepted Answer

$$Statement-2$$ is true since                 ∑$$k=1n$$ $$k2$$−$$(k$$−$$1)3=13$$−$$03+23$$−$$13+$$…                                                            $$+n3$$−$$(n$$−$$1)3=n3$$⇒  ∑$$k=1n$$ $$(k$$−$$(k$$−$$1))k2+k(k$$−$$1)+(k$$−$$1)2=n3$$⇒      ∑$$k=1n$$ $$k2+k(k$$−$$1)+(k$$−$$1)2=n3Putting$$ $$n=20$$, we get            ∑$$k=120$$ $$(k$$−$$1)2+k(k$$−$$1)+k2=203$$⇒            $$1+(1+2+4)+(4+6+9)+$$…$$+$$                   $$(361+380+400)=8000Thus$$, $$Statement-1$$ is also true and $$Statement-2$$ is a correct reason for it.

Statement $- 1$ : The sum of the series $1 + (1 + 2 + 4) + (4 + 6 + 9) + (9 + 12 + 16) + \dots + (361 + 380 + 400)$ is $8000$
Statement $- 2$ : $\sum_{k = 1}^{n} (k^{3} - (k - 1)^{3}) = n^{3}$ for each natural number $n .$

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material

Statement-1: The sum of the series 1+(1+2+4)+(4+6+9)+(9+12+16)+…+(361+380+400) is 8000Statement-2: ∑k=1n k3−(k−1)3=n3 for each natural number n.

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material

Statement $- 1$ : The sum of the series $1 + (1 + 2 + 4) + (4 + 6 + 9) + (9 + 12 + 16) + \dots + (361 + 380 + 400)$ is $8000$
Statement $- 2$ : $\sum_{k = 1}^{n} (k^{3} - (k - 1)^{3}) = n^{3}$ for each natural number $n .$