3if n is even show that n(n2+20 ) is divisible by 48
10 Answers
13 2 = 9 ≡ 1 ( mod 8 )
So , 3 2 n ≡ 1 ( mod 8 )
But , 7 ≡ - 1 ( mod 8 )
Hence , 3 2 n + 7 ≡ - 1 + 1 ( mod 8 ) ≡ 0 ( mod 8 )
So , 3 2 n + 7 is always a multiple of 8 .
1By binomial expansion -
3 2 n + 7
= ( 8 + 1 ) n + 7
= 1 + n C 1 8 + ............( multiples of 8 ) + 7
= 8 + ................multiples of 8
= multiples of 8
3my sol : 32n+7
= (32)n -1+8
now 9n-1n is divisible by (9-1) or 8
therefore 32n+7 is a multiple of 8
39n is what here minch? I thought of proving it by induction..if you put n = 1 you get 21 which is not divisible by 48.
The first one can be done by induction also.
For n = 1, we see that 32 + 7 = 16 is divisible by 8.
Let for n = k, 32k + 7 be a multiple of 8.
So 32k + 7 = 8@ for some natural @.
Now we must prove for n = k + 1.
32(k + 1) + 7
= 32k.32 + 7
= 9(8@ - 7) + 7
= 72@ - 63 + 7
= 72@ - 56
= 8(9@ - 7)
Hence for n = k + 1, the expression is a multiple of 8.
So by Principle of Induction, this holds for all natural numbers.
39Okay, if n is even, substitute n by n = 2@, for some natural @(as n is always even, this condition holds good).
So our expression is 2@(4@² + 20).
For @ = 1, we see that
2(4 + 20)
= 48 is divisible by 48.
Now let for @ = k, our expression be divisible by 48.
2k(4k² + 20)
= 8k3 + 40k is divisible by 48.
So let 8k3 + 40k = 48X, for some natural X.
We must prove for @ = k + 1.
2(k + 1)((2(k + 1))2 + 20)
I tried quite a lot to somehow get 8k3 + 40k into play...but it's not coming right. Can you check if I've done something wrong?
1As DiPrit-H said, take n=2m
Let P(n)= 2m(4m2+20) is divisible by 48.
P(1) is true
We assume P(k)=2k(4k2+20) to be true
Then we have to prove P(k+1) is true
P(k+1)
=2(k+1)[4(k+1)2+20]
=2(k+1)[4k2+8k+4+20]
=2k[4k2+20] +2k(8k+4) + 2[4k2+8k+24]
=48J + 16k2+8k+8k2+16k+48 ......[48J=2k(4k2+20) which is divisible by 48]
=48J + 24k2+24k+48
=48J + 24[k2+k+2]
Now k2+k+2 is always even so 24[k2+k+2] is divisible by 48.
So P(k+1) is true
So, P(n) i.e. 2m(4m2+20]=n(n2+20) is divisible by 48
1solution without induction
since n is even taking n=2k
now it becomes 2k(4k2+20) =8k(k2+5)
inference 1st: term divisible by 8
now taking two cases
a)k is even then k of k(k2+5) term is divisible by 2.
b)k is odd then (k2+5) of k(k2+5) is divisible by 2.
inference 2nd: term divisible by 2*8=16
now writing it as k(k2-1+6) it is divisible by 3(fermat's theorem)
inference 3rd: term divisible by 16*3=48.,
hence term is divisible by 48.
3another method......
n(n2+20)=n(n2 - 4 + 24) = n[(n-2)(n+2)+24]
=[(n-2)n(n+2)+24n]
Now (n - 2) n (n + 2) is the product of three consecutive even integers and
therefore must be divisible by 2 . 4 . 6 or 48 ; also 24n is divisible by 48
therefore n(n2+20) is divisible by 48.
1induction gives the coolest solution,.,.,
observe for n=1,2,3......the identity holds good now...
3^2m+7 is divisible by 8 we r to show 3^(2m+2)+7 is also
3^(2m+2)=9.3^2m=8.3^2m+3^2m
hence proved./././.