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What is the sum of the series 1,2,4,8,16,31.........till 1000 terms?

11 Answers

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The question is somewhat wrong because it must be 1+2+4+8+16+32+....+1000th term

However , now coming to the point 

      1+2+4+8+16+32+..+1000th term

=> 1+(21+22+23+24+....+2999)

=> 1+{2(2999-1)}...This is the correct answer.

 As we all know Sn= Summation of n number digits in a series is= 2(2n-1)

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S=1+2+4+8+16+32+...........

Therefore this is geometric series.

S=1+2+4+8+16+...............+2999

S=(21000-1)/(2-1)

S=21000-1

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The question is somewhat wrong because it must be 1+2+4+8+16+32+....+1000th

      1+2+4+8+16+32+..+1000th term

=> 1+(21+22+23+24+....+2999)

=> 1+{2(2999-1)}... I also work it that the answer 
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Here the sequence 1,2,4,8,16,32 .......till 1000 terms are in the form of geometric progression (GP)

therefore, sum of n terms in GP =  a(r n - 1)/(r - 1) 
                                               = a(1 - r n)/(1 - r) if r<1
Here a= first term
        r = common ratio
        n = no of terms

So here, a = 1
             r = 2
             n =1000

Therefore, sum of series = 1(21000-1)/(2-1) = 21000-1

Therefore sum = 21000-1
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Here the terms 1,2,4,.........(1000 terms)

we can write the above terms as 20 , 21 , 2 , 23 .......2999

The above terms are in GP(geometric progression)

Sum of N terms in GP  = a(rn - 1)/(r-1)     if r>1

Here r = 2/ 2 = 2

a=1  

S1000  = 1(21000 - 1)/(2-1)

S1000  =21000 -1  

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Since this is in GP so and sum in GP is given by-

Sn = a{(rⁿ-1) /(r-1)}
So
Sn = 1 {(2^1000 - 1) /( 2-1) }
Sn = 2^1000 - 1
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The sum of the given series from 1 to 1000 terms is 2047. The pattern of the given series is that the numbers are powers of two, i.e. 2^0, 2^1, 2^2, 2^3, 2^4 and so on. The sum of the series can be calculated using the formula S = (2^n - 1) * 2^(n-1). So, for n = 10, S = (2^10 - 1) * 2^(10-1) = 2047.
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The sum of the series 1,2,4,8,16,31,.... till 1000 terms is 1023. The formula to calculate the sum of an arithmetic progression is given by S = (n/2) [2a + (n - 1) d], where n is the number of terms, a is the first term, and d is the common difference. In this case, n = 1000, a = 1, and d = 2. Therefore, the sum is 1023.
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till 1000 terms are in the form of geometric progression (GP)

therefore, sum of n terms in GP = a(r n - 1)/(r - 1) 

                                               = a(1 - r n)/(1 - r) if r<1

Here a= first term

        r = common ratio

        n = no of terms

So here, a = 1

             r = 2

             n =1000

Therefore, sum of series = 1(21000-1)/(2-1) = 21000-1

Therefore sum = 21000-1
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To find the sum of the given series, we can observe that each term is obtained by doubling the previous term and then subtracting 1. 

Let's break down the series into three parts:

Part 1: 1, 2, 4, 8, 16, ..., which is a geometric series with a common ratio of 2.

Part 2: 31, 63, 127, ..., which is a series obtained by doubling the previous term and then subtracting 1.

Part 3: The last term in the series, which is 1000.

To find the sum of Part 1, we can use the formula for the sum of a geometric series:

Sum = (first term * (1 - common ratio^n)) / (1 - common ratio)

In this case, the first term is 1, the common ratio is 2, and n is the number of terms in Part 1. To find n, we need to solve the equation 2^n = 1000.

Using the formula, we can find the sum of Part 1. Then, we can sum the terms in Part 2 and add the last term of Part 3 to get the final sum of the series.
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If we know the number of terma and the different in these then we use 2 with power of 1000 substrate by 1 this give us the summ of whole series at once. This is real approach 
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