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-1
This series is known as the Harmonic Series and it diverges but very, very slowly. For example, the first 100 terms sum to 5.187...., the first 1000 terms to 7.486...., and the first 1000000 terms to 14.392.... There are many proofs of the divergence of this series and an internet search of Harmonic Series will no doubt find many of them.
This is an arithmetic progression with a first term of 1, a difference of 1, last term (and number of terms) 105. → sum = ½ × number_of_terms × (first_term + last_term) = ½ × 105 × (1 + 105) = ½ × 105 × 106 = 5565
The sum of the numbers given is 31. 1 + 2 + 4 + 8 + ... + 2n = 2n+1 - 1
Sum = 79*(79+1)/2 = 79*80/2 = 3160
-1
There is no solution to the question as asked. If the sum of n terms is 2n+1 then the sum of n+1 terms, using the same formula, is 2*(n+1)+1 = 2n+2+1 = 2n+3 So the (n+1)th term is sum to n+1 minus sum to n = (2n+3) - (2n+1) = 2 So each term is 2. But if each term is 2, then the sum of n terms must be even. The sum is clearly odd - which leads to a contradiction.
Cause its sum 1 who he looks upto!!
the sum of 1 plus 9 is 10 because 1+9
0
This series is known as the Harmonic Series and it diverges but very, very slowly. For example, the first 100 terms sum to 5.187...., the first 1000 terms to 7.486...., and the first 1000000 terms to 14.392.... There are many proofs of the divergence of this series and an internet search of Harmonic Series will no doubt find many of them.
The sum is: 8
let 1+2+4+8+...... =s s= 1+2(1+2+4+8+........) s=1+2s s=-1 but sum of infinite terms cannot be -1 so, what's the answer
The sum is 1511.
This is an arithmetic progression with a first term of 1, a difference of 1, last term (and number of terms) 105. → sum = ½ × number_of_terms × (first_term + last_term) = ½ × 105 × (1 + 105) = ½ × 105 × 106 = 5565
The sum of the numbers given is 31. 1 + 2 + 4 + 8 + ... + 2n = 2n+1 - 1
2