Sequences and Series Test - 19

General term is given by $$(2k)(1+3k)$$ where $$k$$ varies from $$1$$  to $$n-1$$
$$ \displaystyle  \sum_{k=1}^{n-1} (2k)(1+3k) = 2 \sum_{k=1}^{n-1} (k+3k^2)$$.
Substituting the values in the formula (Note that $$n = n-1$$),
We get $$(n)(n-1)(1+(2(n-1)+1))$$$$=2n^3-2n^2$$

$$ x=(-1)^{ { a }^{ 1 } }+(-1)^{ { a }^{ 2 } }+..............+(-1)^{ { a }^{ 1006 } }\\ \because \quad a\quad is\quad not\quad distinct\quad and\quad it\quad is\quad an\quad integer.\\ Let\quad us\quad take\quad a\quad as\quad even.\\ \therefore \quad { a }^{ n }\quad will\quad be\quad even\quad for\quad all\quad values\quad of\quad n.\\ \therefore \quad (-1)^{ { a }^{ n } }\quad will\quad be\quad 1\\ Total\quad number\quad of\quad terms\quad is\quad 1006.\\ \therefore \quad x=\sum _{ n=1 }^{ 1006 }{ (-1)^{ n } } =1006\\ \therefore \quad x=1006\longrightarrow an\quad even\quad number.\\ Again,\quad let\quad a\quad be\quad odd.\quad Then\quad { a }^{ n }\quad will\quad be\quad odd\quad for\quad all\quad \\ values\quad of\quad n.\\ \therefore \quad (-1)^{ { a }^{ n } }=-1\\ \therefore \quad \sum _{ n=1 }^{ 1006 }{ (-1)^{ n } } =\quad -1006\rightarrow an\quad even\quad negative\quad number.\\ Again\quad y=(-1)^{ { a }^{ 1007 } }+(-1)^{ { a }^{ 1008 } }+...................+(-1)^{ { a }^{ 2013 } }\\ As\quad per\quad the\quad previous\quad discussion\\ (-1)^{ { a }^{ n } }=1\quad in\quad case\quad a\quad is\quad even\quad and\quad \\ (-1)^{ { a }^{ n } }=\quad -1\quad in\quad case\quad a\quad is\quad odd\\ Total\quad number\quad of\quad terms=(2013-1007)+1=1007\\ Then\quad y=1007\quad in\quad case\quad a\quad is\quad even\\ \quad \quad \quad \quad \quad =-1007\quad in\quad case\quad a\quad is\quad odd\\ y\quad is\quad an\quad odd\quad number,\quad positive\quad and\quad negative\quad respectively.\\ Let\quad us\quad investigate\quad the\quad options.\\ In\quad case\quad a\quad is\quad odd\longrightarrow \left\{ x\quad is\quad even,\quad negative\\ y\quad is\quad odd,\quad negative \right\} ----(1)\\ In\quad case\quad a\quad is\quad even\longrightarrow \left\{ x\quad is\quad even,\quad positive\\ y\quad is\quad odd,\quad positive \right\} ----(2)                                                                      Option\quad A\longrightarrow (-1)^{ x }=1,\quad (-1)^{ y }=-1\longrightarrow a\quad odd.\\ a\quad even\longrightarrow (-1)^{ x }=1,\quad (-1)^{ y }=-1\\ Option\quad A\quad is\quad not\quad correct.\\ Option\quad B\longrightarrow a\quad is\quad odd\longrightarrow (-1)^{ x }=1,\quad (-1)^{ y }=-1\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad a\quad is\quad even\longrightarrow (-1)^{ x }=1,\quad (-1)^{ y }=-1\\ Option\quad B\quad is\quad correct.\\ Obviously\quad option\quad C\quad and\quad D\quad are\quad not\quad correct.\\ Answer-\quad Option\quad B\quad is\quad correct. $$

unit digit of $$(12357)^{655}$$

Given that

The pattern of numbers are as followed
$$1^2-1=0$$
$$2^2-1=3$$
$$3^2-3=6$$
$$4^2-4=12$$
$$5^2-5=20$$
$$6^2-6=30$$
$$7^2-7=42$$
$$8^2-8=56$$
$$9^2-9=72$$

$$\displaystyle \sum_{n=1}^{10} \int_{0}^{n} x dx =  \sum_{n=1}^{10} \frac{n^{2}}{2} $$

After each term starting with $$p_3$$, the number in from of 4 is twice the number in front of 4 in the preceding term.
Thus the number for the unknown plant is $$2 \times 12 +4 =28$$

If $$n$$ is even,
Let $$n=2m$$
$$\therefore S=1^{3}+3.2^{2}+3^{3}+3.4^{2}+5^{3}+3.6^{2}+....+(2m-1)^{3}+3(2m)^{2}$$

$$=\left \{ 1^{3}+3^{3}+5^{3}+...+(2m-1)^{3} \right \}+3\left \{ 2^{2}+4^{2}+6^{2}+....+(2m)^{2} \right \}$$

$$=\sum_{r=1}^{m}(2r-1)^{3}+3.4\sum_{r=1}^{m}r^{2}$$

$$=\sum_{r=1}^{m}\left \{ 8r^{3}-12r^{2}+6r-1 \right \}+12\sum_{r=1}^{m}r^{2}$$

$$=8\sum_{r=1}^{m}r^{3}-12\sum_{r=1}^{m}r^{2}+6\sum_{r=1}^{m}r-\sum_{r=1}^{m}1+12 \sum_{r=1}^{m}r^{2}$$

$$=8\sum_{r=1}^{m}r^{3}+6\sum_{r=1}^{m}r-\sum_{r=1}^{m}1$$

$$=8.\displaystyle \frac{m^{2}(m+1)^{2}}{4}+6\displaystyle \frac{m(m+1)}{2}-m$$

$$=2m^{2}(m+1)^{2}+3m(m+1)-m$$

$$=m[2m^{3}+4m^{2}+5m+2]$$

$$=\displaystyle \frac{n}{2}\left [ 2\left ( \displaystyle \frac{n}{2} \right )^{3}+4\left ( \displaystyle \frac{n}{2} \right )^{2}+5\left ( \displaystyle \frac{n}{2}\right )+2 \right ]$$  $$(\because m=\displaystyle \frac {n}{2})$$

$$=\displaystyle \frac{n}{8}(n^{3}+4n^{2}+10n+8)$$