Sum of the series 1 - 2 + 3 - 4 + 5 - 6 + ... + (n+2) - (n-1) + n is 2006 then sum of the digits of n is ? Options
1) 3
2) 4
3) 5
4) 6
Correct answer is 6 can you explain
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n=3
1-2+3-4+5-6+7+(n+2)-(n-1)+n
1-2= -1
-1+3= 2
2-4= -2
-2+5= 3
3+6= -3
-3+7= 4
4+(3+2)= 9
9-(3-1)= 7
7+3= 10
answer is = 10
1-2+3-4+5-6+7+(n+2)-(n-1)+n
1-2= -1
-1+3= 2
2-4= -2
-2+5= 3
3+6= -3
-3+7= 4
4+(3+2)= 9
9-(3-1)= 7
7+3= 10
answer is = 10
heerrai24:
Wrong answer answer is 6
Thanks for your answer maybe is there any mistake or you're forgetting something which is causing answer 10 so please check again and help us . I have this McQ question and answer printed where this question answer is 6 but your is 10 I hope you understood
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