Statement 1 The sum of first n terms of the series `1^(2)-2^(2)+3^(2)-4^(2)_5^(2)-"……"` can be `=+-(n(n+1))/(2)`. Statement 2 Sum of first n narural n
Statement 1 The sum of first n terms of the series `1^(2)-2^(2)+3^(2)-4^(2)_5^(2)-"……"` can be `=+-(n(n+1))/(2)`. Statement 2 Sum of first n narural numbers is `(n(n+1))/(2)`
A. Statement 1 is true, Statement 2 is true, Statement 2 is a corrct 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.
1 Answers
Correct Answer - A
Clearly, nth term of the given series is negative or positive according as n is even or odd, respectively.
Case I When n is even, in this case the given series is
`1^(2)-2^(2)+3^(2)-4^(2)+5^(2)-6^(2)+"..."+(n-1)^(2)-n^(2)`
`=(1^(2)-2^(2))+(3^(2)-4^(2))+(5^(2)-6^(2))+"..."+[(n-1)^(2)-n^(2)]`
`=(1-2)(1+2)+(3-4)(3+4)+(5-6)(5+6)+"..."+(n-1-n)(n-1+n)`
`=-(1+2+3+4+5+6+"..."+(n-1)+n)=(n(n+1)/(2))`
Case II When n is odd, in this case the give series is
`1^(2)-2^(2)+3^(2)-4^(2)+5^(2)-6^(2)+"..."+(n-2)^(2)-(n-1)^(2)+n^(2)`
`=(1^(2)-2^(2))+(3^(2)-4^(2))+(5^(2)-6^(2))+"..."+[(n-2)^(2)-(n-1)^(2)]+n^(2)`
`=(1-2)(1+2)+(3-4)(3+4)+(5-6)(5+6)+"..."+[(n-2)-(n-1)][(n-2)+(n-1)]+n^(2)`
`=-[1+2+3+4+5+6+"..."+(n-2)+(n-1)]+n^(2)` ltbr. `=-((n-1)(n-1+1))/(2)+n^(2)=(n(n+1))/(2)`
It is clear that Statement 1 is true, Statement 2 is true, Statement 2 is a corrct explanation for Statement 1.