Question

Assertion :If $S_n{=}^n{C}_1-\frac{{}^n{C}_2}{2}+\frac{{}^n{C}_3}{3}-.....+\frac{{(-1)}^{n-1}{}^n{c}_n}{n}$ & $T_n=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}$, then $S_n=T_n\forall n$ Reason: $S_{n+1}-S_n=T_{n+1}-T_n$

• A. Both Assertion & Reason are individually true & Reason is correct explanation of Assertion,
• B. Both Assertion & Reason are individually true but Reason is not the correct (proper) explanation of Assertion,
• C. Assertion is true but Reason is false,
• D. Assertion is false but Reason is true.

Answer 1

The correct option is A Both Assertion & Reason are individually true & Reason is correct explanation of Assertion,

Since $S_1{=}^1{C}_1=1$ and $T_1=1\Rightarrow S_1=T_1$
Thus if the reason is true and if $S_n=T_n$ for some $n$.

then $S_{n+1}=T_{n+1}\Rightarrow S_n=T_n$ for all $n$
Thus assertion is true if reason is true.
Let us prove reason
$T_{n+1}-T_n=\frac{1}{n+1}$
$S_{n+1}-S_n{=}^{n+1}{C}_1-\frac{{}^{n+1}{C}_2}{2}+\frac{{}^{n+1}{C}_3}{3}-....+\frac{{(-1)}^n{.}^{n+1}{C}_n}{n}$
$+\frac{{(-1)}^n{.}^{n+1}{C}_{n+1}}{n+1}-({}^n{C}_1-\frac{{}^n{C}_2}{3}+\frac{{}^n{C}_3}{3}-.....+\frac{{(-1)}^{n-1}{}^n{C}_n}{n})$
$=\left({}^{n+1}{C}_1{-}^n{C}_1\right)-\frac{1}{2}\left({}^{n+1}{C_2-}^n{C}_1\right)+\frac{1}{3}\left({}^{n+1}{C_3-}^n{C}_3\right)-$
$....+\frac{{(-1)}^{n-1}}{n}\left({}^{n+1}{C}_n{-}^n{C}_n\right)+\frac{{(-1)}^n}{n+1}$
${=}^n{C}_0-\frac{{}^n{C}_1}{2}+\frac{{}^n{C}_2}{3}-.....+\frac{{(-1)}^n{}^n{C}_n}{n+1}$
$={\int }_0^1{(1-x)}^{n}dx=\frac{1}{n+1}$
Thus reason is true.