if the sum of n terms of an AP IS 2n^2+ 5 , then prove that an= 4n + 3
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Given,
S(n) = 2n^2+5.....
then, S(n-1) = 2 ×(n-1)^2 + 5
= 2(n^2-2n+1) +5
= 2n^2-4n+2+5
= 2n^2-4n+7......
We know, a(n) = S(n) - S(n-1)
= 2n^2+5-(2n^2-4n+7)
= 2n^2+5-2n^2+4n-7
= 4n-2.....(H.P.)
IT'S NOT 4n+3, IT'S 4n-2 !!!!
CHECK IT!
S(n) = 2n^2+5.....
then, S(n-1) = 2 ×(n-1)^2 + 5
= 2(n^2-2n+1) +5
= 2n^2-4n+2+5
= 2n^2-4n+7......
We know, a(n) = S(n) - S(n-1)
= 2n^2+5-(2n^2-4n+7)
= 2n^2+5-2n^2+4n-7
= 4n-2.....(H.P.)
IT'S NOT 4n+3, IT'S 4n-2 !!!!
CHECK IT!
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