Sequences and Series

Here $a_n=7+(n-1) 8$ and $\mathrm{T}_1=3$, $a_1=7, d=8$

Also, $ \mathrm{T}_{n+1}=\mathrm{T}_n+a_n$

$\mathrm{T}_n=\mathrm{T}_{n-1}+a_{n-1}$

$\mathrm{T}_2=\mathrm{T}_1+a_1$

$\therefore \mathrm{T}_{n+1}=\left(\mathrm{T}_{n-1}+a_{n-1}\right)+a_n$

$\mathrm{T}_{n+1}=\mathrm{T}_{n-2}+a_{n-2}+a_{n-1}+a_n$

So

$ \begin{aligned} & \mathrm{T}_{n+1}=\mathrm{T}_1+a_1+a_2 \ldots \ldots a_n \\\\ & \mathrm{~T}_{n+1}=\mathrm{T}_1+\frac{n}{2}(2 \times 7+(n-1) 8) \\\\ & \mathrm{T}_{n+1}=\mathrm{T}_1+n(4 n+3) ........(1) \end{aligned} $

For (A), if $ n=19, \mathrm{~T}_{20}=3+(19)(79)=1504$

For (C), if $n=29, \mathrm{~T}_{30}=3+29(119)=3454$

For (B),

$ \begin{aligned} & \sum_{k=1}^{20} \mathrm{~T}_k =\sum_{k=1}^{20}\left(\mathrm{~T}_1+4 n^2+3 n\right)+3 \\\\ & = 3+\sum_{k=1}^{20}\left(3+4 n^2+3 n\right) \\\\ &= 3+3(19)+\frac{3(19)(20)}{2} +\frac{4(19)(20)(34)}{6} \\\\ &= 3+10507=10510 \end{aligned} $

Similarly, for (D)

$ \sum\limits_{k=1}^{30} \mathrm{~T}_k=3+\sum\limits_{k=1}^{29}\left(4 n^2+3 n+3\right)=35615 $

For option (a) ${S_n}(x) = \sum\limits_{k = 1}^n {{{\cot }^{ - 1}}\left[ {{{1 + k(k + 1){x^2}} \over x}} \right]} $ can be written as

${S_n}(x) = \sum\limits_{k = 1}^n {{{\tan }^{ - 1}}\left[ {{{(k + 1)x - kx} \over {1 + kx\,.\,(k + 1)x}}} \right]} $

$ = \sum\limits_{k = 1}^n {[{{\tan }^{ - 1}}(k + 1)x - {{\tan }^{ - 1}}(kx)]} $

$ = {\tan ^{ - 1}}(n + 1)x - {\tan ^{ - 1}}x$

$ = {\tan ^{ - 1}}\left( {{{nx} \over {1 + (n + 1){x^2}}}} \right)$

Now, ${S_{10}}(x) = ta{n^{ - 1}}\left( {{{10x} \over {1 + 11{x^2}}}} \right)$

$ = {\pi \over 2} - {\cot ^{ - 1}}\left( {{{10x} \over {1 + 11{x^2}}}} \right) = {\pi \over 2} - {\tan ^{ - 1}}\left( {{{1 + 11{x^2}} \over {10x}}} \right)$

Option (a) is correct.

For option (b)

$\mathop {\lim }\limits_{n \to \infty } \cot ({S_n}(x)) = \cot \left( {{{\tan }^{ - 1}}\left( {{x \over {{x^2}}}} \right)} \right)$

$ = \cot \left( {{{\tan }^{ - 1}}\left( {{1 \over x}} \right)} \right) = \cot (co{t^{ - 1}}x) = x,\,x > 0$

Option (b) is correct.

For option (c)

${S_3}(x) = {\pi \over 4} \Rightarrow {{3x} \over {1 + 4{x^2}}} = 1$

$ \Rightarrow 4{x^2} - 3x + 1 = 0$ has no real root.

Option (c) is incorrect.

For option (d)

For $x = 1,\,\tan ({S_n}(x)) = {n \over {n + 2}}$ which is greater than ${1 \over 2}$ for n $\ge$ 3

Option (d) is incorrect.

$\begin{aligned} & \text { Given, } \mathrm{S}_n=\sum_{k=1}^{4 n}(-1)^{\frac{k(k+1)}{2}} \cdot \mathrm{K}^2 \\ & \Rightarrow \mathrm{S}_n=-1^2-2^2+3^2+4^2-5^2-6^2+7^2 +8^2-9^2-10^2+11^2+12^2 \ldots \ldots . 4 n \text { terms } \end{aligned}$

$\begin{aligned} \Rightarrow \mathrm{S}_n= & -\left[1^2+5^2+9^2+\ldots . n \text { terms }\right] \\ & -\left[2^2+6^2+10^2+\ldots n \text { terms }\right] \\ & +\left[3^2+7^2+11^2+\ldots n \text { terms }\right] \\ & +\left[4^2+8^2+12^2+\ldots n \text { terms }\right] \end{aligned}$

$\Rightarrow \mathrm{S}_n=-\sum_\limits{r=1}^n(4 r-3)^2-\sum_\limits{r=1}^n(4 r-2)^2 +\sum_\limits{r=1}^n(4 r-1)^2+\sum_\limits{r=1}^n(4 r)^2$

$\begin{aligned} & \Rightarrow \mathrm{S}_n=\sum_{r=1}^n\left((4 r)^2+(4 r-1)^2-(4 r-2)^2-(4 r-3)^2\right) \\ & \Rightarrow \mathrm{S}_n=\sum_{r=1}^n(32 r-12) \\ & \Rightarrow \mathrm{S}_n=32 \sum_{r=1}^n r-\sum_{r=1}^n 12 \\ & \Rightarrow \mathrm{S}_n=32 \cdot \frac{n(n+1)}{2}-12 n \\ & \Rightarrow \mathrm{S}_n=16 n^2+16 n-12 n \\ & \Rightarrow \mathrm{S}_n=4 n(4 n+1) \end{aligned}$

If $n=9$, then $\mathrm{S}_9=1332$

If $n=8$, then $\mathrm{S}_8=1056$

Hints :

(i) Recall $\sum_\limits{\mathrm{K}=1}^{4 n}(-1)^{\frac{k(k+1)}{2}} \cdot \mathrm{K}^2=-\sum_\limits{r=1}^n(4 r-3)^2-\sum_\limits{r=1}^n(4 r-2)^2+\sum_\limits{r=1}^n(4 r-1)^2+\sum_\limits{r=1}^n(4 r)^2$

(ii) $\sum_\limits{r=1}^n a \cdot f(r)+b \cdot g(r)-c \cdot h(r) = \sum_\limits{r=1}^n a \cdot f(r)+\sum_\limits{r=1}^n b \cdot g(r)-\sum_\limits{r=1}^n c \cdot h(r)$

$=a \sum_\limits{r=1}^n f(r)+b \sum_\limits{r=1}^n g(r)-c \sum_\limits{r=1}^n h(r)$

(iii) $\sum_\limits{r=1}^n r=\frac{n(n+1)}{2}, \sum_\limits{r=1}^n a=a r$ where is a constant.

${S_n} < \mathop {\lim }\limits_{x \to \infty } {S_n} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {{1 \over n}{1 \over {1 + {k \over n} + \left( {{k \over {{n^2}}}} \right)}}} $

$ = \int\limits_0^1 {{{dx} \over {1 + x + {x^2}}} = {\pi \over {3\sqrt 3 }}} $

As

$h\sum\limits_{k = 0}^n {f(kh) > \int\limits_0^1 {f(x)dx > h} } $

$\sum\limits_{k = 1}^n {f(kh)} $

So, ${T_n} > {\pi \over {3\sqrt 3 }}$