If `(1+x)^(n) `= `C_(0)+C_(1).x+C_(2). x^(2)+..+C_(n). x^(n) ` then `C_(0)+2. C_(1)+3. C_(2)+..+(n+1). C_(n) `=

Explanation

Here , last term of ` C_(0) + 2C_(1) + 3C_(2) + …+ (n+1) C_(n)` is
`(n +1) C_(n) "i.e."(n=1)` and last term with positive sign.
and ` n + 1 n* 1 + 1`
or`{:("n)n+1(1"),(" "underline(-n)),(" "underline(1)):}`
Here, q = 1 and r = 1
The given series is
`(1 + x)^(n) =C_(0)+ C_(1) x + C_(2) x^(2) + ...+ C_(n)x^(n)`
Now , replacing x by `x^(1)` and multiplying both sides by x , we get
` x(1 + x)^(n)= C_(0) + C_(1) x + C_(2) x^(2)+ ...+ C_(n) x^(n+1) + ...+ (n+1) C_9n) x^(n)`
Putting x = 1 , we get
` n(2)^(n-1)2^(n) = C_(0) + 2C_(1) + 3C_(2) + ...+ (n+1) C_(n)`
Differentiating both sides w.r.t.x, we get
` x* (1 + x)^(n-1) + (1 + x)^(n) *1 = C_(0) + 2C_(1) x + 3C_(2) x^(2) + ...+ (n+1) C_(n) x^(n)`
Putting x = 1 , we get
` n(2)^(n-1) + 2^(n) = C_(0) + 2C_(1) + 3C_(2) + ...+ (n+1) C_(n)`
or ` C_(0) + 2C_(1) + 3C_(2) + ...+ (n+1) C_(n) = (n+2)^(n-1)`
I . Aliter
` LHS = C_(0) + 2C_(1) + 3C_(2) + ...+ (n+1)C_(n)`
` = C_(0) + (1 + 1)C_(1) + (1 +2)C_(2) + ...+ (1 + n)C_(n)`
` = (C_(0) + C_(1) + C_(2) + ...+ C_(n)) + (C_(1) + 2C_(2) + ...+ n C_(n)) " " ` [use example ]
` = 2^(n) + n*2^(n-1) = (n+2)2^(n-1) = RHS `
II .Aliter
LHS ` = C_(0) + 2C_(1) + 3C_(2) + ...+ (n+1) C_(n)`
`= sum_(r=1)^(n+1) r* ""^(n)C_(r-1) = sum_(r=1)^(n+1) (r - 1 + 1)* ""^(n)C_(r-1)`
` = sum_(r=1)^(n+1)(r-1)* ""^(n-1)C_(r)+ ""^(n)C_(r-1)`
` sum_(r=1)^(n+1)n* ""^(n-1)C_(r-2) + sum_(r=1)^(n+1) ""^(n)C_(r-1)`
`[because ""^(n)C_(r-1) = (n)/(r-1) * ""^(n-1)C_(r-2)]`
`=n (0 ""^(n-1)C_(0) + ""^(n-1)C_(1) + ""^(n-1)C_(2) + ...+ ""^(n-1)C_(n-1)) + (""^(n)C_(0) + ""^(n)C_(1) + ""^(n)C_(2) + ...+ ""^(n)C_(n))`
` n* 2^(n-1) + 2^(n) = (n+2)* 2^(n-1) = RHS `