21.
21.
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1 Answer
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21. Let
Putting x = 1 , orP ( 1 ) = x 2 − y 2 is divisible by x + y which is true.( x + y ) ( x − y )isdivisibleby x + y , Assume that P(k) is true for some natural no. k
is divisible by x + yP ( k ) = x 2 k − y 2 k i.e.
wherex 2 k − y 2 k = a ( x + y ) ∈ z (1)⇒ x 2 k = a ( x + y ) + y 2 k Now, let us prove P(k +1) is true.
P ( k + 1 ) : x 2 ( k + 1 ) − y 2 ( k + 1 ) = x 2 x + 2 − y 2 x + 2 = x 2 ⋅ x 2 k − y 2 ⋅ y 2 k = x 2 [ a ( x + y ) + y 2 k ] − y 2 ⋅ y 2 k [using(1) ] = a x 2 ( x + y ) + x 2 ⋅ y 2 k − y 2 ⋅ y 2 k = a x 2 ( x + y ) + y 2 k ( x 2 − y 2 ) = a x 2 ( x + y ) + y 2 k ( x + y ) ( x − y ) = ( x + y ) [ a x 2 + y 2 k ( x − y ) ] = b ( x + y )where b = [ a x 2 + y 2 k ( x − y ) ] ∈ z ⇒ x 2 k + 2 − y 2 k + 2 is divisible by x + y P(k+ 1) is true where P (k) is true.∴ Hence, by P.M.I. P(n) is true for all natural number i.e.
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
Now the numerator again ® 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.
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