Find the modulus of the complex number `z = -3 - i4`

Explanation

Here . Last term of `1^(2) *C_(1) + 2^(2)*C_(2) + 3^(2) *C_(3) + …+ n^(2) *C_(n) ` is
` n^(2)*C_(n) " i.e.,n^(2)` . Linear factor of `n^(2)` are n and, [start
always with greater fector ] and last term with positive sign.
and `{:(n=n*1+0" or " " n)n(1"),(" "underline(-n)),(" "underline(0)):}`
Here ,q= 1 and r=0
Then ,the given series is
`(1+x)^(n) = C_(0) + C_(1) x + C_(2) x^(2) +C_(3)x^(3) + ...+ C_(n)x^(n)`
On differentiating both sides w.r.t.x, we get
`nx(1+x)^(n-1) = C_(1)+ 2C_(2)x + 3C_(3)x^(2) + ...+ n C_(n)x^(n-1) `...(i)
and in last , numberical is n` C_(n) ` i.e. , and power
`(1+x) "is" n-1`
Then ` {:(n=(n-1)*1+0" or " " n)n(1"),(" "underline(-+)),(" "underline(1)):}`
Here .q = 1 and r=1
Now , multiplying both sides by x in Eq .(i) , then
`nx (1+x)^(n-1) = C_(1)x+2C_(2)x^(2) + 3C_(2)x^(3) +...+n^(2)C_(n) x^(n-1)`
Differentiating on both sides w.r.t..x, we get
`n{x*(n-1)(1+x)^(n-2)+(1+x)^(n-1)*1}`
` = C_(1)*1+2^(2)C_(2)x + 3^(2)C_(3)x^(2) + ...+ n^(2)C_(n)x^(n-1)`
Putting x = 1 ,we get
`n{1*(n-1)*2^(n-2)+ 2^(n-1)} = 1^(2)*C_(1) + 2^(3) *C_(2) + 3^(2) *C_(3)+...+n^(2) *C_(n)`
or `1^(2) *C_(1) +2^(2) *C_(2) + 3^(2)*C_(3) +...+ n^(2) *C_(n) = n (n+1)2^(n-2)`
Aliter
`LHS = 1^(2) *C_(1) + 2^(2) *C_(2) + 3^(2)*C_(3) + ...+ n^(2) *C_(n)`
` sum_(r=1)^(n) r^(2)*""^(n)C_(r) = sum_(r=1)^(n) r^(2) *(n)/(r) (""^(n-1)C_(r-1)`
` [because ""^(n)C_(r) = (n)/(r) .""^(n-1)C_(r-1)]`
`= sum_(r=1)^(n) r*""^(n-1)C_(r-1) = n sum_(r=1)^(n) {(r-1)+1}*""^(n-1)C_(r-1)`
` = n sum_(r=1)^(n)(r-1)*""^(n-1)C_(r-1) +n sum_(r=1)^(n) ""^(n-1)C_(r-1)`
` = n sum_(r-1)^(n) (n-1)*""^(n-2)C_(r-2) + n sum_(r=1)^(n) ""^(n-3)C_(r-1)`
`= (n-1)sum_(r=1)^(n) ""^(n-2)C(r-2) + n sum_(r=1)^(n) ""^(n-1)C_(r-1)`
` = n(n-1)(0+""^(n-2)C_(0) + ""^(n-2)C_(1) + ""^(n-2)C_(2) + ...+ ""^(n-2)C_(n-2)) + n(""^(n-1)C_(0) +""^(n-1)C_(1) + ""^(n-1)C_(2) + ...+ ""^(n-1)C_(n-1))`
`= n(n-1) *2^(n-2) + n*2^(n-1) = n (n+1) 2^(n-2) = RHS `