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Prove that 1 * (2^2 - 3^2) + 2 * (3^2 - 4^2) + 3 * (4^2 - 5^2) + ... n terms = n/6 (n + 1) (4n + 11)

Solution:
Your problem `->` Prove that 1 * (2^2 - 3^2) + 2 * (3^2 - 4^2) + 3 * (4^2 - 5^2) + ... n terms = n/6 (n + 1) (4n + 11)


L.H.S. `= 1 × (2^2 - 3^2) + 2 × (3^2 - 4^2) + 3 × (4^2 - 5^2) + ... n` terms

`= sum [ f(n) ]`

`= sum [ n ((n + 1)^2 - (n + 2)^2) ]`

`= sum [ n (n^2 + 2n + 1 - n^2 - 2n - 4) ]`

`= sum [ n (-2n - 3) ]`

`= sum (-2n^2 - 3n) ]`

`= -2 sum n^2 - 3 sum n`

`= -2 * (n (n + 1) (2n + 1))/6 - 3 * (n (n + 1))/2`

`= - n/6 (n + 1) [ 2(2n + 1) + 9 ]`

`= - n/6 (n + 1) (4n + 11)`

`=` R.H.S. (Proved)






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