Let `S_n=sum_(k=1)^(4n)(-1)(k(k+1))/2k^2dot` Then `S_n` can take value (s) `1056` b. `1088` c. `1120` d. `1332`

Options

A. 1056
B. 1088
C. 1120
D. 1332

Answer

A, D

Explanation

Convert it into differences and use sum of n terms of an AP,
i.e, `S_(n) = (n)/(2) [2a + (n -1) d]`
Now, `S_(n) = underset(k =1)overset(4n)sum (-1)^((k(k+1))/(2)).k^(2)`
`= -(1)^(2) - 2^(2) + 3^(2) + 4^(2) - 5^(2) - 6^(2) + 7^(2) + 8^(2) +`....
`= (3^(2) -1^(2)) + (4^(2) -2^(2)) + (7^(2) -5^(2)) + (8^(2) -6^(2)) +`...
`ubrace(2"{"(4 + 6 + 12+ ...))_("n terms")+ubrace((6+14+22+...)"}")_("n terms")`
`= 2[(n)/(2) {2xx4 + (n-1) 8} + (n)/(2) {2 xx 6 + (n -1) 8}]`
`= 2 [n (4 + 4n - 4) + n (6 + 4n -4)]`
`= 2 [4n^(2) + 4n^(2) + 2n] = 4n (4n +1)`
Here, `1056 = 32 xx 33, 1088 = 32 xx 34`,
`1120 = 32 xx 35, 1332 = 36 xx 37`
1056 and 1332 are possible answer