Question

If $n>3$ and $a,bϵR,$ then the value of $ab-n(a-1)(b-1)+\frac{n(n-1)}{1.2}(a-2)(b-2)-......+(-1{)}^n(a-n\left)\right(b-n)$ is equal to

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

$T_{k+1}$ = $(-1{)}^k$ ${}^{.n}{C}_k(a-k)(b-k)$
$=(-1{)}^k.$ ${}^n{C}_k[ab-k(a+b)+k^2]$
Thus, the sum of series in (i)
${\sum }_{k-0}^n(-1{)}^k{.}^n{C}_k[ab-k(a+b)+k^2]$
$=ab{\sum }_{k-0}^n(-1{)}^k{.}^n{C}_k-(a+b\left){\sum }_{k-0}^n\right(-1{)}^{k}k{.}^n{C}_k+{\sum }_{k-0}^n(-1{)}^k{k}^2{.}^n{C}_k$
We know that
$(1+x{)}^n{=}^n{C}_0{+}^n{C}_{1}x{+}^n{C}_2{x}^2+...{+}^n{C}_n{x}^n...(ii)$
Differentiating both sides w.r.t. x, we get
$n(1+x{)}^{n-1}={1.}^n{C}_1+{2.}^n{C}_{2}x+{3.}^n{C}_3{x}^2+...+n{.}^n{C}_n{x}^{n-1}...(iii)$
Multiplying it by x we get
$n.x(1+x{)}^{n-1}={1.}^n{C}_{1}x+{2.}^n{C}_2{x}^2+{3.}^n{C}_3{x}^3+...+n{.}^n{C}_n{x}_n$
Differentiating w.r.t. x, we get
$n(1+x{)}^{n-1}+n(n-1\left)x\right(1+x{)}^{n-2}$
$=1^2{.}^n{C}_1+2^2{.}^n{C}_{2}x+...+n^2{.}^n{C}_n{x}^{n-1}...\left(iv\right)$
Putting x = –1 in (ii), (iii) and (iv), we get
${\sum }_{k-0}^n(-1{)}^k{.}^n{C}_k=0$
${\sum }_{k-0}^n(-1{)}^{k}k{.}^n{C}_k=0$
${\sum }_{k-0}^n(-1{)}^k{k}^2{.}^n{C}_k=0$
Thus, the sum of the series is 0