Binomial Theorem Test - 58

The general term in the expansion of $$(5^{\tfrac 12} + 7^{\tfrac 18})^{1024}$$ is
$$T_{r+1} = ^{1024}C_r (5^{\tfrac 12})^{1024-r} (7^{\tfrac 18})^r$$

$$= ^{1024}C_r 5^{512-\tfrac{r}{2}} 7^{\tfrac{r}{8}}$$

$$ = (^{1024}C_r 5^{512-r})(5^{\tfrac{r}{2}}7^{\tfrac{r}{8} })$$

$$= (^{1024}C_r 5^{512r}) (5^4\times 7)^{\tfrac{r}{8}}$$

Clearly $$T_{r+1}$$ will be an integer, if $$\dfrac{r}{8}$$ is an integer such that $$0\leq r\leq 1024$$
$$\Rightarrow r$$ is a multiple of $$8$$ lying satisfying $$0\leq r \leq 1024$$
$$ r = 0, 8, 16, ... 1024$$
$$\Rightarrow r$$ can assume $$129$$ values.
Hence, there are $$129$$ integral terms in the expansion of $$(5^{\tfrac 12} + 7^{\tfrac 18})^{1024}$$

$$S=$$ coefficient of $${x}^{r}$$ in $$E$$ where $$E={ (1+x) }^{ n }+3{ (1+x) }^{ n-1 }+5{ (1+x) }^{ n-2 }+...+(2(n-r)+1)({ (1+x) }^{ r }$$
Let $$F={ a }^{ n }+3{ a }^{ n-1 }+.....(2(n-r)+1){ a }^{ r }$$
$$\Rightarrow \cfrac { 1 }{ a } F={ a }^{ n-1 }+....+(2(n-r)-3){ a }^{ r }+(2(n-r)+1){ a }^{ r-1 }\quad $$
$$\Rightarrow \left( 1-\cfrac { 1 }{ a }  \right) F={ a }^{ n }+2{ a }^{ n-1 }+....+2{ a }^{ r }-(2(n-r)+1){ a }^{ r-1 }\quad $$
$$={ a }^{ n }+\cfrac { 2{ a }^{ r }(1-{ a }^{ n-r }) }{ 1-a } -(2(n-r)+1){ a }^{ r-1 }$$
$$\Rightarrow F=\cfrac { { a }^{ n+1 } }{ a-1 } -\cfrac { 2{ a }^{ r+1 }(1-{ a }^{ n-r }) }{ { (1-a) }^{ 2 } } -\cfrac { (2(n-r)+1){ a }^{ r } }{ a-1 } $$
Thus,
$$\quad E=\cfrac { { (1+x) }^{ n+1 } }{ x } +\cfrac { 2{ (1+x) }^{ n+1 } }{ { x }^{ 2 } } -\cfrac { 2{ (1+x) }^{ r+1 } }{ { x }^{ 2 } } -\cfrac { (2(n-r)+1){ (1+x) }^{ r } }{ x } $$
$$\therefore \quad S={ _{  }^{ n+1 }{ C } }_{ r+1 }+2({ _{  }^{ n+1 }{ C } }_{ r+2 })\quad ={ _{  }^{ n+2 }{ C } }_{ r+2 }+({ _{  }^{ n+1 }{ C } }_{ r+2 })$$
Hence, option B.