Prove that `2. 7^n+3. 5^n-5`is divisible by 24, for all `n in N`.

1.	Prove that `2. 7^n+3. 5^n-5`is divisible by 24, for all `n in N`.
Answer» Given expression is. `27^n+35^n-5` When `n = 1`, given expression is, `27+35-5 = 12+15-5 = 24` So, for `n=1`, given expession is divisible by `24`. Let for any `k in N`, given expression is divisible by `24`. Then, `27^k+35^k-5 = 24c`, where `c` is a natural number. `=>35^k = 24c-27^k+5->(1)` Now, we have to prove, for `n = k+1`, given expression is divisible by `24`. For `n = k+1`, given expression is, `27^(k+1)+35^(k+1) - 5` `=147^k+535^k - 5` From (1), given expression becomes, `=147^k+5(24c-27^k+5) -5` `=147^k+120c-107^k+25-5` `=47^k+120c+20` `=4*7^k+120c+24-4` `=4(7^k -1)+24(5c+1)` `=4((1+6)^k -1)+ 24(5c+1)` `=4(C(k,0)6^k+C(k,1)6^(k-1)+C(k,2)6^(k-2)+...C(k,k)-1)+ 24(5c+1) ``=4(C(k,0)6^k+C(k,1)6^(k-1)+C(k,2)6^(k-2)+..+C(k,k-1)6^1.+1-1)+ 24(5c+1) ``=4(C(k,0)6^k+C(k,1)6^(k-1)+C(k,2)6^(k-2)+...6)+ 24(5c+1) ``(C(k,0)6^k+C(k,1)6^(k-1)+C(k,2)6^(k-2)+...6)` is divisible by `6`, so we can replace this term with `6a`, where `a` is another non-zero fixed number. `= 24a+24(5c+1)` `=24(a+5c+1)`, which is clearly divisible by `24`. Thus, given expression will be divisible by `24`.

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