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

(B) Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1

(A) Statement 1 is false, statement 2 is true

(C) Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1

(D) Statement 1 is true, statement 2 is false

JEE Main · 2012 · 29 Apr 2012 · Offline · Sequences and Series

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 any natural number n.

(A)Statement 1 is false, statement 2 is true
(B)Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1✓ Correct
(C)Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1
(D)Statement 1 is true, statement 2 is false

Answer: (B)

Solution

The identity in the second statement is true:

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

telescopes, every interior term cancelling, to leave

n^{3}

. Looking at the series in the first statement, its

k

-th bracket is

(k - 1)^{2} + k (k - 1) + k^{2}

, which is exactly

k^{3} - (k - 1)^{3}

. Summing the brackets for

k = 1

up to

k = 20

therefore collapses to

2 0^{3} = 8000

, so the first statement is true as well. Because the telescoping identity is the very mechanism that produces the sum, it is a correct explanation of the first statement.

JEE Main · 2012 · 29 Apr 2012 · Offline · Single correct option

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JEE Main · 2012 · 29 Apr 2012 · Offline · Sequences and Series

Solution

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