The sum of seven consecutive natural numbers, the first of which is n

Answer 1

Since this is an arithmetic series we use S = n ( I1 + I2) / 2 , where n = number of terms, I1 is first term, I2 is last. For this problem n = 7 and the last number is n + 6. Plug in to get S = 7 (n + n + 6) /2 which reduces to 7 (n + 3)

Answer 2

And now that you know how to use a variable in your calculations, try for the sum of:

1 + (1/4) + (1/9) + (1/16) + (1/25) + ..... out to infinity. (i.e. sum of the reciprocals of the squares of the positive natural numbers). You can check on Google, just enter the first few terms.

-- Cheers, -- Mr. d.

Answer 3

The seven consecutive number are n , n + 1, n+2, n+3, n+4, n+5, n+6.

Sum = n + (n+1) + (n+2 )+ (n+3) + (n+4) +(n+5) + (n+6) = 7n + 21

Answer 4

n + (n+1) + (n+2) + (n+3) + (n+4) + (n+5) + (n+6) = 7n+21 = 7(n+3)