2021/106) Evaluate closed form of $1^2+2^2+3^2+\cdots+n^2$

We have

$(k+1)^3 = k^ 3 + 3k ^2 + 3k + 1$

Or $(k+1)^3 - k^ 3 = 3k ^2 + 3k + 1$

Adding from 1 to n we get

$\sum_{k=1}^n((k+1)^3 - k^ 3) = 3\sum_{k=1}^n k^2 + 3\sum_{k=1}^nk + \sum_{k=1}^n1$

The LHS is a telespcopic sum = $(n+1)^3-1$

We know $\sum_{k=1}^nk = \frac{n(n+1)}{2}$

so we get $(n+1)^3 - 1 = 3\sum_{k=1}^n k^2 + 3\frac{n(n+1)}{2} + n$

or  $3\sum_{k=1}^n k^2 = (n+1)^3 - 1 -  3\frac{n(n+1)}{2} - n$

$=n^3 + 3n^2 + 3n + 1 - 3\frac{n^2+n}{2} - n$

$= \frac{1}{2}(2n^3 + 6n^2 + 6n - 3n^2 -3n -n )$

$= \frac{1}{2}(2n^3 + 3n^2 + 2n )$

$=\frac{1}{2}n(2n^2+ 3n + 2)$

$= \frac{1}{2}n(n+2)(2n+1)$

so $\sum_{k=1}^n k^2 = \frac{1}{6}(n(n+2)(2n+1)$

2021/105) Prove that if k = mn and k is a perfect square and m and n are co-primes them m and n are perfect squares

 Now let p be a prime factor of k.

So p is a prime factor of m or n but not both because GCD(m, n) = 1

Now because k is a square p shall occur even number of times say 2m

All the 2m occurences must be factor of m (as we have mention p is factor of m) 

So any prime factor of k whcich is a factor of m shall occur even number of times in m and which is not a factor of m shall occur even number of times in n making both m and n perfect squares.

2021/104) Prove that for positive integer n we have $n^2 | (n+1)^n-1$

We have

$(n+1)^n - 1$

$= \sum_{k=0}^{n}  {n \choose k} n^{n-k} -1 $

$= \sum_{k=0}^{n-2}  {n \choose k} n^{n-k} + {n \choose n-1} n^{n-(n-1)} + {n \choose n} n^{n-n} -1$

$= \sum_{k=0}^{n-2}  {n \choose k} n^{n-k} +n * n + 1 - 1$

$= \sum_{k=0}^{n-2}  {n \choose k} n^{n-k} +n^2$

now each term in sum is having $n^2$ as a term and hence the expression is divisible by $n^2$

2021/103) Show that there are infinite numbers of the form $10^n+3$ that are composite

Because the number is not divsible by 2,3,5 for any n so let us check if is divisible by 7 for some n. 

Now as 7 is co-prime to 10 so as per fermats little theorem 

$10^6 \equiv 1 \pmod 7\cdots(1)$ 

By checking from1 to 6 we see that 

$3^4 = 81 \equiv  4 \pmod 7$

or  $10^4 = 81 \equiv  4 \pmod 7$ as $ 10\equiv 3 \pmod 7$

using (1) we get 

$10^{6k+ 4} = \equiv  4 \pmod 7$

or  $10^{6k+ 4} + 3 = \equiv  0 \pmod 7$

hence divisible by 7 and hence composite

so there are infinite numbers as k goes from 1 onwards are composite for n = 6k + 4

2021/102) What is the largest natural number n below 50 such that LCM (n, n + 1, . . . , 50) = LCM (1, 2, . . . , 50), where LCM stands for least common multiple.?

 it is 27. because if we leave out 27 then none of the numbers above 27 is divisible by 27 so we shall not have a factor 27 of the LCM but for numbers 1 to 50 LCM shall have a factor 27.

27 is not by magic. it has to a prime number or a power of a prime. if it is composite other than power of a prime then is has got smaller factor and this is taken care of because smaller numbers are taken care of. so we should look for a number greater than 1/2 of the number because if a is taken care of the 2a is taken care of. in LCM. so we look through numbers 26,27 etc and find 27.

2021/101) Show that any number > 6 can be written as sum of 2 co-prime numbers

Let us consider 3 cases. Let the number be n 

1) n is odd then 2 numbers  are (2, n-1)

2) n is even and of the form 4n. Then the 2 numbers (2n-1, 2n+1)

3) n is even and of the form 4n +2. Then the 2 numbers (2n -1 , 2n + 3)  

2021/100) prove: $\sum_{k=0}^{n} \binom{n}{k}^2 =\dfrac{(2n)!}{(n!)^2}$

Out of 2n objects n objects can be chosen in

$\dfrac{(2n)!}{(n!)^2}$ ways

Now let us make 2n objects into 2 groups of n objects each.

For picking n objects from the set we need k objects from 1st set and n-k from 2nd set and n varies from 0 to n so number of ways

$\sum_{k=0}^{n} \binom{n}{k} \binom{n}{n-k}$

The 2 above are same as it shows the number of ways in 2 different ways

So

$\dfrac{(2n)!}{(n!)^2}= \sum_{k=0}^{n} \binom{n}{k} \binom{n}{n-k}$

Now as $\binom{n}{k} = \binom{n}{n-k}$ so we get the result