6.1.2 + 2.3 + 3.4 + … + n (n + 1) =
6.1.2 + 2.3 + 3.4 + … + n (n + 1) =
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1 Answer
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6. Let the given statement be P (n) i.e.,
P (n)=1.2+2.3+3.4+ … +2 (n+1)=
For n=1,
P (1)=1.2=2= = =2.
Which is true.
considerP (k) be true for some positive integer k
1.2 + 2.3 + 3.4 + … + k (k + 1) = - (1)
Now, let us prove that P (k+1) is true.
Here, 1.2 + 2.3 + 3.4 + … + k (k + 1) + (k+1) (k+2)
By using (1), we get
=
= (k+1) (k+2)
=
By further simplification;
P (k+1) is true whenever P (k) is true.
Therefore, by the principle of mathematical induction, statement P (n) is true for all natural no. i.e., n.
Similar Questions for you
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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