Question

PROVE THAT 1000 = 2000
Please follow me .......
Plzz give the correct answer or else don't answer​

Answer 1

Answer:

yes

Step-by-step explanation:

We know that 1+2+…+n=n(n+1)2. So we can write

m+(m+1)+(m+2)+…+n=(1+2+…+n)−(1+2+…+(m−1))=n(n+1)2−(m−1)m2=n2+n−m2+m2=(m+n)(n−m+1)2.

Find the sum of all the integers between 1000 and 2000 which are not divisible by 5.

We can calculate this by summing all the integers between 1000 and 2000, and then subtracting the sum of all those divisible by 5. By the above formula,

∑i=10002000i=(1000+2000)(2000−1000+1)2=3000×10012=1501500.

The sum of all numbers between 1000 and 2000 that are divisible by 5 can be written as

1000+1005+⋯+1995+2000=5×200+5×201+⋯+5×400=5×(200+201+⋯+400)=5∑i=200400i=5×(200+400)×(400−200+1)2=5×600×2012=301500.

Hence, the desired sum is

1501500−301500=1200000.

A geometric series has first term 2 and common ratio 0.95. The sum of the first n terms of the series is denoted by Sn and the sum to infinity is denoted by S. Calculate the least value of n for which S−Sn<1.

We have

Sn=a1−rn1−r,

where a=2 and r=0.95, and

S=limn→∞Sn=a1−r=20.05=40.

We are to find the least value of n for which S−Sn<1; let’s try to solve S−Sn=1.

This gives us 40−21−0.95n1−0.95=40−40(1−0.95n)=1, and multiplying out and rearranging gives

0.95n=140⟹n=ln(1/40)ln0.95=71.9...

Thus the value of n that we require is 72.

Alternatively, we could solve 0.95n=140 by taking logs to base 0.95.