Sequences and Series
Let $S = 2 + {6 \over 7} + {{12} \over {{7^2}}} + {{20} \over {{7^3}}} + {{30} \over {{7^4}}} + \,.....$. Then 4S is equal to
If a1, a2, a3 ...... and b1, b2, b3 ....... are A.P., and a1 = 2, a10 = 3, a1b1 = 1 = a10b10, then a4 b4 is equal to -
$x = \sum\limits_{n = 0}^\infty {{a^n},y = \sum\limits_{n = 0}^\infty {{b^n},z = \sum\limits_{n = 0}^\infty {{c^n}} } } $, where a, b, c are in A.P. and |a| < 1, |b| < 1, |c| < 1, abc $\ne$ 0, then :
If $A = \sum\limits_{n = 1}^\infty {{1 \over {{{\left( {3 + {{( - 1)}^n}} \right)}^n}}}} $ and $B = \sum\limits_{n = 1}^\infty {{{{{( - 1)}^n}} \over {{{\left( {3 + {{( - 1)}^n}} \right)}^n}}}} $, then ${A \over B}$ is equal to :
The sum 1 + 2 . 3 + 3 . 32 + ......... + 10 . 39 is equal to :
Let x, y > 0. If x3y2 = 215, then the least value of 3x + 2y is
If $\{ {a_i}\} _{i = 1}^n$, where n is an even integer, is an arithmetic progression with common difference 1, and $\sum\limits_{i = 1}^n {{a_i} = 192} ,\,\sum\limits_{i = 1}^{n/2} {{a_{2i}} = 120} $, then n is equal to :
Let $a_{1}, a_{2}, a_{3}, \ldots$ be an A.P. If $\sum\limits_{r=1}^{\infty} \frac{a_{r}}{2^{r}}=4$, then $4 a_{2}$ is equal to _________.
Explanation:
Given
$S = {{{a_1}} \over 2} + {{{a_2}} \over {{2^2}}} + {{{a_3}} \over {{2^3}}} + {{{a_4}} \over {{2^4}}}\, + \,.....\,\infty $
${{{1 \over 2}S = {{{a_1}} \over {{2^2}}} + {{{a_2}} \over {{2^3}}}\, + \,.........\,\infty } \over {{S \over 2} = {{{a_1}} \over 2} + {{({a_2} + {a_1})} \over {{2^2}}} + {{({a_3} + {a_2})} \over {{2^3}}}\, + \,......\,\infty }}$
$ \Rightarrow {S \over 2} = {{{a_1}} \over 2} + {d \over 2}$
$ \Rightarrow {a_1} + d = {a_2} = 4 \Rightarrow 4{a_2} = 16$
If $\frac{1}{2 \times 3 \times 4}+\frac{1}{3 \times 4 \times 5}+\frac{1}{4 \times 5 \times 6}+\ldots+\frac{1}{100 \times 101 \times 102}=\frac{\mathrm{k}}{101}$, then 34 k is equal to _________.
Explanation:
$S = {1 \over {2 \times 3 \times 4}} + {1 \over {3 \times 4 \times 5}} + {1 \over {4 \times 5 \times 6}}\, + \,....\, + \,{1 \over {100 \times 101 \times 102}}$
$ = {1 \over {(3 - 1)\,.\,1}}\left[ {{1 \over {2 \times 3}} - {1 \over {101 \times 102}}} \right]$
$ = {1 \over 2}\left( {{1 \over 6} - {1 \over {101 \times 102}}} \right)$
$ = {{143} \over {102 \times 101}} = {k \over {101}}$
$\therefore$ $34k = 286$
Explanation:
${1 \over {{3^{12}}}} + 5\left( {{{{2^0}} \over {{3^{12}}}} + {{{2^1}} \over {{3^{11}}}} + {{{2^2}} \over {{3^{10}}}}\, + \,.......\, + \,{{{2^{11}}} \over 3}} \right) = {2^n}\,.\,m$
$ \Rightarrow {1 \over {{3^{12}}}} + 5\left( {{1 \over {{3^{12}}}}{{\left( {{{(6)}^2} - 1} \right)} \over {(6 - 1)}}} \right) = {2^n}\,.\,m$
$ \Rightarrow {1 \over {{3^{12}}}} + {5 \over 5}\left( {{1 \over {{3^{12}}}}\,.\,{2^{12}}\,.\,{3^{12}} - {1 \over {{3^{12}}}}} \right) = {2^n}\,.\,m$
$ \Rightarrow {1 \over {{3^{12}}}} + {2^{12}} - {1 \over {{3^{12}}}} = {2^n}\,.\,m$
$ \Rightarrow {2^n}\,.\,m = {2^{12}}$
$ \Rightarrow m = 1$ and $n = 12$
$m\,.\,n = 12$
$ \frac{2^{3}-1^{3}}{1 \times 7}+\frac{4^{3}-3^{3}+2^{3}-1^{3}}{2 \times 11}+\frac{6^{3}-5^{3}+4^{3}-3^{3}+2^{3}-1^{3}}{3 \times 15}+\cdots+ \frac{30^{3}-29^{3}+28^{3}-27^{3}+\ldots+2^{3}-1^{3}}{15 \times 63}$ is equal to _____________.
Explanation:
${T_n} = {{\sum\limits_{k = 1}^n {\left[ {{{(2k)}^3} - {{(2k - 1)}^3}} \right]} } \over {n(4n + 3)}}$
$ = {{\sum\limits_{k = 1}^n {4{k^2} + {{(2k - 1)}^2} + 2k(2k - 1)} } \over {n(4n + 3)}}$
$ = {{\sum\limits_{k = 1}^n {(12{k^2} - 6k + 1)} } \over {n(4n + 3)}}$
$ = {{2n(2{n^2} + 3n + 1) - 3{n^2} - 3n + n} \over {n(4n + 3)}}$
$ = {{{n^2}(4n + 3)} \over {n(4n + 3)}} = n$
$\therefore$ ${T_n} = n$
${S_n} = \sum\limits_{n = 1}^{15} {{T_n} = {{15 \times 16} \over 2} = 120} $
If $\sum\limits_{k=1}^{10} \frac{k}{k^{4}+k^{2}+1}=\frac{m}{n}$, where m and n are co-prime, then $m+n$ is equal to _____________.
Explanation:
$\sum\limits_{k = 1}^{10} {{k \over {{k^4} + {k^2} + 1}}} $
$ = {1 \over 2}\left[ {\sum\limits_{k = 1}^{10} {\left( {{1 \over {{k^2} - k + 1}} - {1 \over {{k^2} + k + 1}}} \right)} } \right.$
$ = {1 \over 2}\left[ {1 - {1 \over 3} + {1 \over 3} - {1 \over 7} + {1 \over 7} - {1 \over {13}}\, + \,...\, + \,{1 \over {91}} - {1 \over {111}}} \right]$
$ = {1 \over 2}\left[ {1 - {1 \over {111}}} \right] = {{110} \over {2\,.\,111}} = {{55} \over {111}} = {m \over n}$
$\therefore$ $m + n = 55 + 111 = 166$
Different A.P.'s are constructed with the first term 100, the last term 199, and integral common differences. The sum of the common differences of all such A.P.'s having at least 3 terms and at most 33 terms is ___________.
Explanation:
${d_1} = {{199 - 100} \over 2} \notin I$
${d_2} = {{199 - 100} \over 3} = 33$
${d_3} = {{199 - 100} \over 4} \notin I$
${d_n} = {{199 - 100} \over {i + 1}} \in I$
${d_i} = 33 + 11,\,9$
Sum of CD's $ = 33 + 11 + 9$
$ = 53$
The series of positive multiples of 3 is divided into sets : $\{3\},\{6,9,12\},\{15,18,21,24,27\}, \ldots$ Then the sum of the elements in the $11^{\text {th }}$ set is equal to ____________.
Explanation:
Given series
$\therefore$ 11th set will have $1 + (10)2 = 21$ term
Also upto 10th set total $3 \times k$ type terms will be $1 + 3 + 5\, + \,......\, + \,19 = 100 - $ term
$\therefore$ Set $11 = \{ 3 \times 101,\,3 \times 102,\,......\,3 \times 121\} $
$\therefore$ Sum of elements $ = 3 \times (101 + 102\, + \,...\, + \,121)$
$ = {{3 \times 222 \times 21} \over 2} = 6993$
Let $a, b$ be two non-zero real numbers. If $p$ and $r$ are the roots of the equation $x^{2}-8 \mathrm{a} x+2 \mathrm{a}=0$ and $\mathrm{q}$ and s are the roots of the equation $x^{2}+12 \mathrm{~b} x+6 \mathrm{~b}=0$, such that $\frac{1}{\mathrm{p}}, \frac{1}{\mathrm{q}}, \frac{1}{\mathrm{r}}, \frac{1}{\mathrm{~s}}$ are in A.P., then $\mathrm{a}^{-1}-\mathrm{b}^{-1}$ is equal to _____________.
Explanation:
Let $\frac{1}{p}, \frac{1}{q}, \frac{1}{r}, \frac{1}{s}$ as $\alpha-3 \beta, \alpha-\beta, \alpha+\beta, \alpha+3 \beta$
So sum of roots $2 \alpha-2 \beta=4$ and $2 \alpha+2 \beta=-2$
Clearly $\alpha=\frac{1}{2}$ and $\beta=-\frac{3}{2}$
Now product of roots, $\frac{1}{p} \cdot \frac{1}{r}=\frac{1}{2 a}=-5 \Rightarrow \frac{1}{a}=-10$
and $\frac{1}{q} \cdot \frac{1}{x}=\frac{1}{6 b}=-8 \Rightarrow \frac{1}{b}=-48$
So, $\frac{1}{a}-\frac{1}{b}=38$
Let $a_{1}=b_{1}=1, a_{n}=a_{n-1}+2$ and $b_{n}=a_{n}+b_{n-1}$ for every
natural number $n \geqslant 2$. Then $\sum\limits_{n = 1}^{15} {{a_n}.{b_n}} $ is equal to ___________.
Explanation:
Given,
${a_n} = {a_{n - 1}} + 2$
$ \Rightarrow {a_n} - {a_{n - 1}} = 2$
$\therefore$ In this series between any two consecutives terms difference is 2. So this is an A.P. with common difference 2.
Also given ${a_1} = 1$
$\therefore$ Series is = 1, 3, 5, 7 ......
$\therefore$ ${a_n} = 1 + (n - 1)2 = 2n - 1$
Also ${b_n} = {a_n} + {b_{n - 1}}$
When $n = 2$ then
${b_2} - {b_1} = {a_2} = 3$
$ \Rightarrow {b_2} - 1 = 3$ [Given ${b_1} = 1$]
$ \Rightarrow {b_2} = 4$
When $n = 3$ then
${b_3} - {b_2} = {a_3}$
$ \Rightarrow {b_3} - 4 = 5$
$ \Rightarrow {b_3} = 9$
$\therefore$ Series is = 1, 4, 9 ......
= 12, 22, 32 ....... n2
$\therefore$ ${b_n} = {n^2}$
Now, $\sum\limits_{n = 1}^{15} {\left( {{a_n}\,.\,{b_n}} \right)} $
$ = \sum\limits_{n = 1}^{15} {\left[ {(2n - 1){n^2}} \right]} $
$ = \sum\limits_{n = 1}^{15} {2{n^3} - \sum\limits_{n = 1}^{15} {{n^2}} } $
$ = 2\left( {{1^3} + {2^3} + \,\,...\,\,{{15}^3}} \right) - \left( {{1^2} + {2^2} + \,\,...\,\,{{15}^2}} \right)$
$ = 2 \times {\left( {{{15 \times 16} \over 2}} \right)^2} - \left( {{{15(16) \times 31} \over 6}} \right)$
$ = 27560$
Let for $f(x) = {a_0}{x^2} + {a_1}x + {a_2},\,f'(0) = 1$ and $f'(1) = 0$. If a0, a1, a2 are in an arithmatico-geometric progression, whose corresponding A.P. has common difference 1 and corresponding G.P. has common ratio 2, then f(4) is equal to _____________.
Explanation:
Given,
$f(x) = {a_0}{x^2} + {a_1}x + {a_2}$
$f'(0) = 1$
$f'(1) = 0$
a0, a1, a2 are in A. G. P
Common difference of $AP = 1$
Common ratio of $GP = 2$
A.P terms = a, a + 1, a + 2
G.P terms = y, ry, r2y
$\therefore$ AGP terms = ay, (a+1)ry, (a+2)r2y
$\therefore$ ${a_0} = ay$
${a_1} = (a + 1)ry = (a + 1)2y$
${a_2} = (a + 2){r^2}y = (a + 2)4y$
Now, $f'(x) = 2x{a_0} + {a_1}$
$\therefore$ $f'(0) = {a_1} = 1$
and $f'(1) = 2{a_0} + {a_1} = 0$
$ \Rightarrow 2{a_0} + 1 = 0$
$ \Rightarrow {a_0} = - {1 \over 2}$
$\therefore$ $ay = - {1 \over 2}$
and $(a + 1)2y = 1$
$ \Rightarrow 2ay + 2y = 1$
$ \Rightarrow 2 \times \left( { - {1 \over 2}} \right) + 2y = 1$
$ \Rightarrow 2y = + \,2$
$ \Rightarrow y = + \,1$
$\therefore$ $a = - {1 \over 2}$
$\therefore$ ${a_2} = (a + 2)4y$
$ = \left( { - {1 \over 2} + 2} \right) \times 4\,.\,1$
$ = 6$
$\therefore$ $f(x) = - {1 \over 2}{x^2} + x + 6$
$\therefore$ $f(4) = - {1 \over 2}{(4)^2} + 4 + 6$
$ = - 8 + 10$
$ = 2$
Let 3, 6, 9, 12, ....... upto 78 terms and 5, 9, 13, 17, ...... upto 59 terms be two series. Then, the sum of the terms common to both the series is equal to ________.
Explanation:
1st AP :
3, 6, 9, 12, ....... upto 78 terms
t78 = 3 + (78 $-$ 1)3
= 3 + 77 $\times$ 3
= 234
2nd AP :
5, 9, 13, 17, ...... upto 59 terms
t59 = 5 + (59 $-$ 1)4
= 5 + 58 $\times$ 4
= 237
Common term's AP :
First term = 9
Common difference of first AP = 3
And common difference of second AP = 4
$\therefore$ Common difference of common terms
AP = LCM (3, 4) = 12
$\therefore$ New AP = 9, 21, 33, .......
tn = 9 + (n $-$ 1)12 $\le$ 234
$ \Rightarrow n \le {{237} \over {12}}$
$ \Rightarrow n = 19$
$\therefore$ ${S_{19}} = {{19} \over 2}\left[ {2.9 + (19 - 1)12} \right]$
$ = 19(9 + 108)$
$ = 2223$
Let for n = 1, 2, ......, 50, Sn be the sum of the infinite geometric progression whose first term is n2 and whose common ratio is ${1 \over {{{(n + 1)}^2}}}$. Then the value of
${1 \over {26}} + \sum\limits_{n = 1}^{50} {\left( {{S_n} + {2 \over {n + 1}} - n - 1} \right)} $ is equal to ___________.
Explanation:
${S_n} = {{{n^2}} \over {1 - {1 \over {{{(n + 1)}^2}}}}} = {{n{{(n + 1)}^2}} \over {n + 2}} = ({n^2} + 1) - {2 \over {n + 2}}$
Now ${1 \over {26}} + \sum\limits_{n = 1}^{50} {\left( {{S_n} + {2 \over {n + 1}} - n - 1} \right)} $
$ = {1 \over {26}} + \sum\limits_{n = 1}^{50} {\left\{ {({n^2} - n) + 2\left( {{1 \over {n + 1}} - {1 \over {n + 2}}} \right)} \right\}} $
$ = {1 \over {26}} + {{50 \times 51 \times 101} \over 6} - {{50 \times 51} \over 2} + 2\left( {{1 \over 2} - {1 \over {52}}} \right)$
$ = 1 + 25 \times 17(101 - 3)$
$ = 41651$
Let A = {1, a1, a2 ....... a18, 77} be a set of integers with 1 < a1 < a2 < ....... < a18 < 77.
Let the set A + A = {x + y : x, y $\in$ A} contain exactly 39 elements. Then, the value of a1 + a2 + ...... + a18 is equal to _____________.
Explanation:
It means all other sums are already present in these 39 values, which is only possible in case when all numbers are in A.P.
Let the common difference be '$d$'.
$77=1+19 \mathrm{~d} \Rightarrow d=4$
So, $\sum\limits_{i=1}^{18} a_{1}=\frac{18}{2}\left[2 a_{1}+17 d\right]=9[10+68]=702$
If the sum of the first ten terms of the series
${1 \over 5} + {2 \over {65}} + {3 \over {325}} + {4 \over {1025}} + {5 \over {2501}} + \,\,....$
is ${m \over n}$, where m and n are co-prime numbers, then m + n is equal to ______________.
Explanation:
${T_r} = {r \over {{{(2{r^2})}^2} + 1}}$
$ = {r \over {{{(2{r^2} + 1)}^2} - {{(2r)}^2}}}$
$ = {1 \over 4}{{4r} \over {(2{r^2} + 2r + 1)(2{r^2} - 2r + 1)}}$
${S_{10}} = {1 \over 4}\sum\limits_{r = 1}^{10} {\left( {{1 \over {(2{r^2} - 2r + 1)}} - {1 \over {(2{r^2} + 2r + 1)}}} \right)} $
$ = {1 \over 4}\left[ {1 - {1 \over 5} + {1 \over 5} - {1 \over {13}} + \,\,....\,\, + \,\,{1 \over {181}} - {1 \over {221}}} \right]$
$ \Rightarrow {S_{10}} = {1 \over 4}\,.\,{{220} \over {221}} = {{55} \over {221}} = {m \over n}$
$\therefore$ $m + n = 276$
If a1 (> 0), a2, a3, a4, a5 are in a G.P., a2 + a4 = 2a3 + 1 and 3a2 + a3 = 2a4, then a2 + a4 + 2a5 is equal to ___________.
Explanation:
Let G.P. be a1 = a, a2 = ar, a3 = ar2, .........
$\because$ 3a2 + a3 = 2a4
$\Rightarrow$ 3ar + ar2 = 2ar3
$\Rightarrow$ 2ar2 $-$ r $-$ 3 = 0
$\therefore$ r = $-$1 or ${3 \over 2}$
$\because$ a1 = a > 0 then r $\ne$ $-$1
Now, a2 + a4 = 2a3 + 1
ar + ar3 = 2ar2 + 1
$a\left( {{3 \over 2} + {{27} \over 8} - {9 \over 2}} \right) = 1$
$\therefore$ a = ${8 \over 3}$
$\therefore$ a2 + a4 + 2a5 = a(r + r3 + 2r4)
$ = {8 \over 3}\left( {{3 \over 2} + {{27} \over 8} + {{81} \over 8}} \right) = 40$
For a natural number n, let ${\alpha _n} = {19^n} - {12^n}$. Then, the value of ${{31{\alpha _9} - {\alpha _{10}}} \over {57{\alpha _8}}}$ is ___________.
Explanation:
${\alpha _n} = {19^n} - {12^n}$
Let equation of roots 12 & 19 i.e.
${x^2} - 31x + 228 = 0$
$ \Rightarrow (31 - x) = {{228} \over x}$ (where x can be 19 or 12)
$\therefore$ ${{31{\alpha _9} - {\alpha _{10}}} \over {57{\alpha _8}}} = {{31({{19}^9} - {{12}^9}) - ({{19}^{10}} - {{12}^{10}})} \over {57({{19}^8} - {{12}^8})}}$
$ = {{{{19}^9}(31 - 19) - {{12}^9}(31 - 12)} \over {57({{19}^8} - {{12}^8})}}$
$ = {{228({{19}^8} - {{12}^8})} \over {57({{19}^8} - {{12}^8})}} = 4$.
The greatest integer less than or equal to the sum of first 100 terms of the sequence ${1 \over 3},{5 \over 9},{{19} \over {27}},{{65} \over {81}},$ ...... is equal to ___________.
Explanation:
$S = {1 \over 3} + {5 \over 9} + {{19} \over {27}} + {{65} \over {81}}\, + $ ....
$ = \sum\limits_{r = 1}^{100} {\left( {{{{3^r} - {2^r}} \over {{3^r}}}} \right)} $
$ = 100 - {2 \over 3}{{\left( {1 - {{\left( {{2 \over 3}} \right)}^{100}}} \right)} \over {1/3}}$
$ = 98 + 2{\left( {{2 \over 3}} \right)^{100}}$
$\therefore$ $[S] = 98$
Explanation:
Given,
${l_1},{l_2},\,.......,\,{l_{100}}$ are in A.P with common difference ${d_1}$.
So from property of A.P we can say,
${l_2} = {l_1} + {d_1}$
${l_3} = {l_1} + 2{d_1}$
$ \vdots $
${l_{100}} = {l_1} + 99{d_1}$
Also given,
${w_1},{w_2},\,......\,,\,{w_{100}}$ are in A.P with common difference ${d_2}$.
$\therefore$ From the property of A.P we can say,
${w_2} = {w_1} + {d_2}$
${w_3} = {w_1} + 2{d_2}$
$ \vdots $
${w_{100}} = {w_1} + 99{d_2}$
Now, also given,
${d_1}{d_2} = 10$
and ${R_i}$ is a rectangle whose length is ${l_i}$ and width is ${w_i}$ and area ${A_i}$.
$\therefore$ We know, area of rectangle
${A_i} = {l_i} \times {w_i}$
$\therefore$ ${A_{51}} = {l_{51}} \times {w_{51}}$
and ${A_{50}} = {l_{50}} \times {w_{50}}$
Given, ${A_{51}} - {A_{50}} = 1000$
$ \Rightarrow ({l_1} + 50{d_1})({w_1} + 50{d_2}) - ({l_1} + 49{d_1})({w_1} + 49{d_2}) = 1000$
$ \Rightarrow [{l_1}{w_1} + 2500{d_1}{d_2} + 50({l_1}{d_2} + {d_1}{w_1})] - [{l_1}{w_1} + 49 \times 49{d_1}{d_2} + 49({l_1}{d_2} + {w_1}{d_1})] = 1000$
$ \Rightarrow [{(50)^2}{d_1}{d_2} - {(49)^2}{d_1}{d_2}] + (50 - 49)({l_1}{d_2} + {d_1}{w_1}) = 1000$
$ \Rightarrow (99 \times 1){d_1}{d_2} + {l_1}{d_2} + {d_1}{w_1} = 1000$
$ \Rightarrow 99 \times 10 + {l_1}{d_2} + {w_1}{d_1} = 1000$
$ \Rightarrow {l_1}{d_2} + {w_1}{d_1} = 10$
Now,
${A_{100}} - {A_{90}}$
$ = {l_{100}}\,.\,{w_{100}} - {l_{90}}\,.\,{w_{90}}$
$ = ({l_1} + 99{d_1})({w_1} + 99{d_2}) - ({l_1} + 89{d_1})({w_1} + 89{d_2})$
$ = [{l_1}{w_1} + {(99)^2}{d_1}{d_2} + 99({l_1}{d_2} + {w_1}{d_1})] - [{l_1}{w_1} + {(89)^2}{d_1}{d_2} + 89({l_1}{d_2} + {w_1}{d_1})]$
$ = [{(99)^2} - {89^2}]{d_1}{d_2} + 10({l_1}{d_2} + {w_1}{d_1})$
$ = 188 \times 10 \times {d_1}{d_2} + 10 \times 10$
$ = 188 \times 10 \times 10 + 100$
$ = 18800 + 100$
$ = 18900$
Let $a_{1}, a_{2}, a_{3}, \ldots$ be an arithmetic progression with $a_{1}=7$ and common difference 8. Let $T_{1}, T_{2}, T_{3}, \ldots$ be such that $T_{1}=3$ and $T_{n+1}-T_{n}=a_{n}$ for $n \geq 1$. Then, which of the following is/are TRUE ?
If $\alpha, \beta, \gamma$ are the roots of the equation $3 x^3-26 x^2+52 x-24=0$ such that $\alpha, \beta, \gamma$ are in geometric progression and $\alpha<\beta<\gamma$, then $3 \alpha+2 \beta+\gamma=$
$68 / 3$
$56 / 3$
12
24
Suppose that the three points $A, B$ and $C$ in the plane are such that their $x$-coordinates as well as $y$-coordinates are in GP with the same common ratio. Then, the points $A, B$ and $C$
Let a1, a2, ...... a40 be in AP and h1, h2, ..... h10 be in HP. If a1 = h1 = 2 and a10 = h10 = 3, then a4h7 is
Let a1, a2, a3 .... be a harmonic progression with a1 = 5 and a20 = 25. The least positive integer n for which an < 0, is
In a sequence of 21 terms, the first 11 terms are in AP with common difference 2 and the last 11 terms are in GP with common ratio 2. If the middle term of AP be equal to the middle term of the GP, then the middle term of the entire sequence is
The sum $\sum\limits_{n = 4}^\infty {\left( {{{2{S_n}} \over {n!}} - {1 \over {(n - 2)!}}} \right)} $ is equal to :
${3 \over {{1^2} \times {2^2}}} + {5 \over {{2^2} \times {3^2}}} + {7 \over {{3^2} \times {4^2}}} + ....$ is :
and ${{2 + 4 + 6 + .... + 2y} \over {3 + 6 + 9 + ..... + 3y}} = {4 \over {{{\log }_{10}}x}}$, then the ordered pair (x, y) is equal to :
${1 \over {x + 1}} + {2 \over {{x^2} + 1}} + {{{2^2}} \over {{x^4} + 1}} + ...... + {{{2^{100}}} \over {{x^{{2^{100}}}} + 1}}$ when x = 2 is :
100$\alpha$ $-$ 199$\beta$ = (100)(100) + (99)(101) + (98)(102) + ...... + (1)(199), then the slope of the line passing through ($\alpha$, $\beta$) and origin is :
$\sum\limits_{n = 1}^\infty {{{{n^2} + 6n + 10} \over {(2n + 1)!}}} $ is equal to :
$1 + {2 \over 3} + {7 \over {{3^2}}} + {{12} \over {{3^3}}} + {{17} \over {{3^4}}} + {{22} \over {{3^5}}} + ......$ is equal to :