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Find the sum of the series {2+3+5+9+8+15+11+21+...} to (2n+1) terms Ans. Rearranging the above terms, we get, S={(2+5+8+11+...(2n+1)/2 terms)+(3+9+15+21+...(2n+1)/2 terms)} Therefore , S=(2n+1)/4 [ 4+ 3{ (2n+1)/2 - 1 } ] + (2n+1)/4 [ 6+ 6{ (2n+1)/2 - 1 } ] Which evaluates to (36n 2 +40n-11)/8 But the answer is (9n 2 +7n+4)/2 Please Help

Find the sum of the series {2+3+5+9+8+15+11+21+...} to (2n+1) terms
 
Ans.
Rearranging the above terms, we get,
S={(2+5+8+11+...(2n+1)/2 terms)+(3+9+15+21+...(2n+1)/2 terms)}
 Therefore ,
 S=(2n+1)/4 [ 4+ 3{ (2n+1)/2 - 1 } ]  +  (2n+1)/4 [ 6+ 6{ (2n+1)/2 - 1 } ]
Which evaluates to
(36n2+40n-11)/8
 
But the answer is
 
(9n2+7n+4)/2
 
Please Help

Grade:12th pass

1 Answers

Latika Leekha
askIITians Faculty 165 Points
8 years ago
Hello student,
The method that you have opted for in this question is incorrect. Please look at the following solution:
The given series is {2+3+5+9+8+15+11+21+...} to (2n+1) terms
Rearranging the terms in the series we have,
S = { 2+5+8+11+14+...} + {3+9+15+21+....}
Since, total number of terms in the series are odd (2n+1), so as divided above, the first sseries has (n+1) terms and the second has n terms.
Let SA = {2+5+8+11+14+... …. (n+1)terms}
SB = {3+9+15+21+... … n terms}
Hence, our series is S = SA + SB
Now, SA = {2+5+8+11+14+... …. (n+1)terms}
a = 2, d = 3 , N= n+1
So, l = a+(N-1)d
= 2+3n
SA = N/2 (a+l)
= (3n2 + 7n + 4)/2 ….…... (*)
SB = {3+9+15+21+... … n terms}
a = 3, d = 6, N = n
So, l = a + (N-1)d
= 3+(n-1)6
= 6n-3
SB = n/2 (3+6n-3)
= 3n2 ….….….….….. (**)
So, S2n+1 = SA + SB
= (3n2 + 7n + 4)/2 + 3n2
= (9n2 + 7n + 4)/2

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