Question

if A=2^65 and B=2^64+2^63+...+2^1+2^0 and c=2^64+2^63+...+2^2+2^1 prove that 2A=B+C+3

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Proved that 2A=B+C+3.

Given:

A=$2^{65}$

B= $2^6^4+2^6^3+...+2^2+2^1$ + 1

c= $2^6^4+2^6^3+...+2^2+2^1$

To find:

Prove that  2A=B+C+3

Explanation:

To prove above equation we can compare left hand side and right hand side.

Left hand side, = 2A

                          =2×$2^{65}$ = $2^{66}$

Right hand side = B+C+3

                          = ( $2^6^4+2^6^3+...+2^2+2^1$)+( $2^6^4+2^6^3+...+2^2+2^1$+ 1) + 3

                          = 2× ( $2^6^4+2^6^3+...+2^2+2^1$) +1+ 3

                          =$2^6^5+2^6^4+...+2^2+2^1$ =4

         $2^6^5+2^6^4+...+2^2+2^1$ This is geometric progression.

Sum of GP = $\frac{a (r^{n-1})}{n-1}$   a = first term = 2

                                  r = Common ratio = 2

                                  n = no. of terms = 65

Sum of GP =  $\frac{a (r^{n-1})}{r-1}$  = $\frac{2 (2^{65-1})}{2-1}$

                 = $2^{66}$ -4

            Right hand side = 2× ( $2^6^4+2^6^3+...+2^2+2^1$) +4

                                       =  $2^{66}$ -4 + 4

                                       =  $2^{66}$

Left hand side =  Right hand side

Hence proved...

To learn more.....

1) In an infinite geometric progression,the sum of first two terms is 6 and every term is four times the sum of all the terms that follow it.find 1. The geometric progression 2.its sum to infinity. Plz help

https://brainly.in/question/38168

2 For each geometric progression find the common ratio ‘r’, and then find an(i) 3, 3/2, 3/4, 3/8, ......... (ii) 2, −6, 18, −54(iii) −1, −3, −9, −18 .... (iv) 5, 2, 4/5, 8/25, .........

https://brainly.in/question/5483880