Statement-1: The sum of the series 1+(1+2+4)+(4+6+9)+(9+12+16)++ (361+380+400) is 8000 . Statement-2: _{k=1}ⁿ (k³-(k-1)³) = n³, for any natural number n. [AIEEE 2012]

Correct Answer is : Statement- 1 is true, Statement- 2 is true; Statement- 2 is a correct explanation for Statement-1.

Statement- 1 is false, Statement- 2 is true.

Statement- 1 is true, Statement- 2 is true; Statement- 2 is a correct explanation for Statement-1.

Statement- 1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.

Statement- 1 is true, Statement- 2 is false.

Correct Answer is : Statement- 1 is true, Statement- 2 is true; Statement- 2 is a correct explanation for Statement-1.

Test Index

AIEEE 2012 Solved Paper

Section: Mathematics

Question : 66 of 89

Marks: +1, -0

1+(1+2+4)+(4+6+9)+(9+12+16)+\dots+

(361+380+400)

\sum\limits_{k=1}^{n} (k^3-(k-1)^3) = n^3

n

Statement- 1 is false, Statement- 2 is true.
Statement- 1 is true, Statement- 2 is true; Statement- 2 is a correct explanation for Statement-1.
Statement- 1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
Statement- 1 is true, Statement- 2 is false.

Solution:

n^{\;\text{th}\;}

term of the given series

\;=T_n=(n-1)^2+(n-1)n+n^2

\;=\;\frac{(n-1)^3-n^3}{(n-1)-n}

\;=n^3-(n-1)^3

\;\Rightarrow S_n=\sum\limits_{k=1}^{n} [k^3-(k-1)^3]

\;\Rightarrow 8000=n^3

\Rightarrow n=20

which is a natural number.

Now, put

n=1,2,3,\dots,20

T_1=1^3-0^3

T_2=2^3-1^3

T_{20}=20^3-19^3

Now,

T_1+T_2+\dots+T_{20}=S_{20}

\Rightarrow S_{20}=20^3-0^3=8000

Hence, both the given statements are true and statement 2 supports statement 1 .

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