CBSE Class 10 Answered
hoe to solve
Asked by Afreen Jilani | 09 Jun, 2013, 03:17: PM
Expert Answer
This can be proved using induction.
So step 1 would be to assume that n=1
Hence, n^2-1 = 1-1 = 0 which is divisible by 8
Step 2 would be to assume k is odd and k^2-1 is divisible by 8
Now, odd numbers differ by 2, so we have to prove
(k+2)^2-1 is divisible by 8
(k+2)^2 - 1 = k^2 +4k +4 - 1 = (k^2-1)+4(k+1)
Since, we have already assumed that k^2-1 is divisible by 8
and since k is odd, so k+1 is even => 4(k+1) is divisible by 8
sum of above 2 terms is also divisible by 8
=> (k+2)^2-1 is divisible
Hence n^2-1 is divisible by 8 for all odd positive integer numbers.
Now, odd numbers differ by 2, so we have to prove
(k+2)^2-1 is divisible by 8
(k+2)^2 - 1 = k^2 +4k +4 - 1 = (k^2-1)+4(k+1)
Since, we have already assumed that k^2-1 is divisible by 8
and since k is odd, so k+1 is even => 4(k+1) is divisible by 8
sum of above 2 terms is also divisible by 8
=> (k+2)^2-1 is divisible
Hence n^2-1 is divisible by 8 for all odd positive integer numbers.
Answered by | 09 Jun, 2013, 07:39: PM
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