Statement 1: The sum of the series `1""+""(1""+""2""+""4)""+""(4""+""6""+""9)""+""(9""+""12""+""16)""+"". . . . . .` `+""(361""+""380""+""400)"" i s ""8000` . Statement 2: `sum_(k=1)^n(k^3-(k-1)^3)=n^3` for any natural number n. (1) Statement 1 is false, statement 2 is true (2) Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1 (3) Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1 (4) Statement 1 is true, statement 2 is false

Options

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

Answer

A

Explanation

`:.(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.