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Statement 1 The sum of the series 1+(1+...

Statement 1 The sum of the series `1+(1+2+4)+(4+6+9)+(9+12+16)+"……."+(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 true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1

B

Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1

C

Statement 1 is true, Statement 2 is false

D

Statement 1 is false, Statement 2 is true

Text Solution

Verified by Experts

The correct Answer is:
A

`:.(1)=(1-0)(1^(2)+1*0+2^(2))=1^(3)-0^(3)`
`(1+2+4)=(2-1)(2^(2)+2*1+1^(2))=2^(3)-1^(3)`
`(4+6+9)=(3-2)(3^(2)+3*2+2^(2))=3^(3)-2^(3)`
`" " vdots " " vdots " "vdots" "`
`(361+380+400)=(20-19)(20^(2)+20*19+19^(2))=20^(3)-19^(3)`
Required sum
`=(1^(3)-0^(3))+(2^(3)-1^(3))+(3^(3)-2^(3))+"......"+(20^(3)-19^(3))=20^(3)=8000`
Also, `sum_(k=1)^(n)k^(3)-(k-1)^(3)=sum_(k=1)^(n){k-(k-1)}{k^(2)+k(k-1)+(k-1)^(2)}`
`=sum_(k=1)^(n)(3k^(2)-3k+1)=3sumn^(2)-3sumn+sum1`
`=(3n(n+1)(2n+1))/(6)-(3n(n+1))/(2)+n`
`=(n)/(2)(2n^(2)+3n+1-3n-3+2)=n^(3)`
Both statement are correct and Statement 2 is the correct explanation of Statement 1.
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