Question

prove that 3^(n)-2n^(2)-1 is divisible by 8?​

Answer 1

Yes it is divisible by 8.

Given:

$3^n-2n^2-1$

To find: We have to prove that it is divisible by 8.

Solution:

The expression is-

$3^n-2n^2-1$

Now let n=1

Putting the value of n as 1 we get-

$3^n-2n^2-1 \\ = {3}^{1} - 2 \times {1}^{2} - 1 \\ = 0$

Thus 0 is divisible by 8.

Again if we take n is equal to 3 then putting the value of n in the expression we get-

$3^n-2n^2-1 \\ {3}^{3} - 2 \times {3}^{2} - 1 \\ 27 - 18 - 1 \\ = 8$

Thus 8 is also divisible by 8.

So, we can say that for any value of n the expression is divisible by 8.

Answer 2

Proved that  3ⁿ - 2n² - 1 is divisible by 8

Step 1:

Assume that P(n) = 3ⁿ - 2n² - 1

Step 2:

Substitute n = 1

P(1) = 3 - 2(1)² - 1

P(1) = 3 - 2 - 1  

P(1) = 0

Hence Divisible by 8

True for n = 1

Step 3:

Assume that for n = k  , p(n) is divisible by 8

P(k) = $3^k-2k^2-1=8m$  

$3^k=8m+2k^2+1$  

Step 4:

Check for  n = k+1  ,

$P(k+1)=3^{k+1}-2(k+1)^2-1$

$3.3^{k}-2(k^2+2k+1)-1$

Substitute $3^k=8m+2k^2+1$  

$3.(8m+2k^2+1)-2k^2 -4k - 3$

$24m+6k^2+3-2k^2 -4k - 3$

$24m+4k^2 -4k$

$24m+4k(k-1)$

k and k -1 are consecutive number hence product must be even number

24m + 4(2q)

= 8(3m + q)

Hence divisible by 8

P(k+1) is true if P(k) is true

Hence Proved using mathematical induction

that  3ⁿ - 2n² - 1 is divisible by 8