practice problems

Q1. Can product of eight consecutive integers be the perfect 4-th power of an integer?
S1.
n(n+1)(n+2)...(n+7)
rearrange:
n(n+7)(n+1)(n+6)(n+2)(n+5)(n+3)(n+4)
Let a = n^2 + 7n + 6
Then product becomes:
P = (a-6).a.(a+4).(a+6) = a^4 + 4a^3 -36a^2 - 144a
(a+1)^4 = a^4 + 4a^3 + 6a^2 + 4a + 1
This is clearly more than the above product. Why? Since a is positive.
(a+1)^4 > P
Can we show a^4 < P?
P = a^4 + 4a(a^2 - 9a - 36), a^2 - 9a - 36 = (a-12)(a+3) > 0  since a >= 14 for n = 1
=> a^4 < P < (a+1)^4 H.P. that it can't be 4th power.

Q2.

S2.
Evaluate the expression:
Denominator would become: k(k+1)
Numerator = k^4 + 2k^3 + 3k^2 + 2k + 1 = (k^2+k+1)^2
So each term is:

(k^2 + k + 1)/(k^2 + k) = 1 + 1/k(k+1)
1/k(k+1) = 1/k - 1/k+1
Now upon addition:

1s add upto 40.
And telescopic sum for the rest:
1/1 - 1/2
1/2 - 1/3
.....
1/40 - 1/41
Add all to get: 1/1 - 1/41 = 40/41
So the final sum is 40 + 40/41
a + b = 80 = answer

Another method:
if x + y + z = 0 then:
(1/x + 1/y + 1/z)^2 = 1/x^2 + 1/y^2 + 1/z^2

Since
1 + k -(k+1) = 0 so:
1/1^2 + 1/k^2 + 1/(-(k+1))^2 = (1 + 1/k + 1/(k+1))^2

Q3.

S3.
= [(k+1)^2 - k^2]/[k^2.(k+1)^2] = 1/k^2 - 1/(k+1)^2
Telescopic sum => 
1/1^2 - 1/(N+1)^2 = 0.9999
=> 1/(N+1)^2 = 0.0001
=> N+1 = 100 => N = 99 = answer