answer this question fast plz it's urgent 26&28 plz
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26)
Let P(n) be the statement:
" a + ( a + d ) + ( a + 2d ) + ... + ( a + (n-1)d ) = (n/2) ( 2a + (n-1)d ) "
We need to:
a) Show that P(1) is true.
b) Show that if P(n) is true, then P(n+1) is true.
a) Putting n = 1, P(1) is the statement:
" a = (1/2) ( 2a ) "
The LHS is "a" and the RHS simplifies also to "a", so this is true.
b) Suppose P(n) is true. ( This is called the inductive hypothesis. )
Then
LHS of P(n+1)
= a + ( a + d ) + ( a + 2d ) + ... + ( a + nd )
= a + ( a + d ) + ( a + 2d ) + ... + ( a + (n-1)d ) + ( a + nd )
= (n/2) ( 2a + (n-1)d ) + ( a + nd ) [ using the inductive hypothesis ]
= (1/2) ( 2na + n(n-1)d + 2a + 2nd )
= (1/2) ( 2(n+1)a + (n²-n+2n)d )
= (1/2) ( 2(n+1)a + (n²+n)d )
= (1/2) ( 2(n+1)a + n(n+1)d )
= ( (n+1)/2 ) ( 2a + nd )
= RHS of P(n+1)
So, using the assumption that P(n) is true, we have shown that P(n+1) would also be true.
With (a) and (b) combined, it follows by the principle of mathematical induction that P(n) is true for all n = 1, 2, 3, 4, ...
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28) x = 2, y = 3, n = 5
By the binomial expansion and the given values, the 2nd, 3rd and 4th terms are:
n x^(n-1) y = 240 (A)
n(n-1)/2 x^(n-2) y² = 720 (B)
n(n-1)(n-2)/6 x^(n-3) y³ = 1080 (C)
Doing (B) divided by (A) we get:
(n-1)/2 y/x = 720/240 = 3 => x/y = (n-1)/6 (D)
Doing (C) divided by (B) we get:
(n-2)/3 y/x = 1080/720 = 3/2 => x/y = 2(n-2)/9 (E)
Putting (D) and (E) together, we have
(n-1)/6 = 2(n-2)/9
=> 3(n-1) = 4(n-2)
=> 3n - 3 = 4n - 8
=> n = 8 - 3 = 5 (*)
Putting this into (D) gives
x/y = 4/6 = 2/3 => y = 3x/2.
Putting this and n=5 into (A) gives:
5 x⁴ y = 240 => x⁴ y = 48 => x⁴ ( 3x / 2 ) = 48
=> x⁵ = 96/3 = 32
=> x = 2
Finally,
y = 3x/2 = 3.