19. n(n+1)(n +5) is a multiple of 3.
19. n(n+1)(n +5) is a multiple of 3.
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1 Answer
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19. We can write the given statement as
P (n): n (n +1) (n+5), which is multiple of 3.
If n= 1, we get
P (1)=1 (1+1) (1+5)=12, which is a multiple of 3 which is true.
Consider P (k) be true for some positive integer k
k (k+1) (k+ 5) is a multiple of 3
k (k+1) (k+5)= 3 m, where (1)
Now, let us prove that P (k + 1) is true
Here,
(k+ 1) { (k+1)+ 1} { (k+1)+ 5}
We can write it as
= (k +1) (k+ 2) { (k + 5) + 1}
By Multiplying the terms.
By eqn. (1)
= 3m + 2 (k + 1) (k + 5) + (k + 1) (k + 2)
= 3m + (k + 1) {2 (k + 5) + (k +2)}
= 3m + (k + 1) {2k + 10 +k + 2}
= 3m + (k + 1) (3k +12)
= 3m + 3 (k + 1) (k+ 4)
=3 {m + (k + 1) (k + 4)}
3 9 wh
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Let base = b
For this limit to be defined 2x3 – 7x2 + ax + b should also trend to 0 or x ® 1.
2 – 7 + (a + b) = 0
(a + b) = 5 …………….(i)
Now this becomes % form we apply L’lopital rule
Now the numerator again ® 0 as x = 1
6x2 – 14x + a ® 0 as x = 1
6 . (1)2 – 14 + a = 0
a = 8 …………….(ii)
a + b = 5
(b = -3) ® from (i) & (ii)
So
When x = 0, y = 0 gives
So, for x = 2, y = 12
24. Let P (n) be the statement “ 2n+7< (n+3)2”
ofn=1
P (1): 2
9<16 which is true. This P (1) is true.
Suppose P (k) is true.
P (k)= 2k+7< (k+3)2 . (1)
Lets prove that P (k +1) is also true.
“ 2 (k + 1) + 7 < (k + 4)2=k2+ 8k + 16”
P (k +1) = 2 (k +1) +7 = (2k +7) +2
< (k +3)2+ 2 (Using 1)
= k2+ 9 + 6k +2 = k2+6k +11
Adding and subtracting (2k + k) in the R. H. S.
for all
is true.
By the principle of mathematical induction, P (n)is true for all n N.
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