For positive integer n, define f(n)=n+(16+5 n-3 n2/4 n+3 n2)+(32+n-3 n2/8 n+3 n2)+(48-3 n-3 n2/12 n+3 n2)+ ldots+(25 n-7 n2/7 n2) Then, the value of displaystyle lim n arrow ∞ f(n) is equal to

Question

For positive integer n, define f(n)=n+(16+5 n-3 n2/4 n+3 n2)+(32+n-3 n2/8 n+3 n2)+(48-3 n-3 n2/12 n+3 n2)+ ldots+(25 n-7 n2/7 n2) Then, the value of displaystyle lim n arrow ∞ f(n) is equal to (A) 3+(4/3) log e 7 (B) 4-(3/4) log e((7/3)) (C) 4-(4/3) log e((7/3)) (D) 3+(3/4) log e 7

Tardigrade Admin · Accepted Answer

f(n)=n+ displaystyle∑r=1n (16 r+(9-4 r) n-3 n2/4 r n+3 n2) f(n)=n+ displaystyle∑r=1n ((16 r+9 n)-(4 r n+3 n2)/4 r n+3 n2) f(n)=n+( displaystyle∑r=1n (16 r+9 n/4 r n+3 n2))-n displaystyle lim n arrow ∞ f(n)= displaystyle lim n arrow ∞ ∑ (16 r+9 n/4 r n+3 n2) = displaystyle lim n arrow ∞ ∑r=1n ((16((r/n))+9) (1/n)/4((r)n)+3) =∫ limits01 (16 x+9/4 x+3) d x=∫ limits01 4 d x-∫ limits01 (3 d x/4 x+3) =4-(3/4)(ℓ n |4 x+3|)01 =4-(3/4) ℓ operatornamen (7/3)

Q. For positive integer $n$ , define
$f (n) = n + \frac{16 + 5 n - 3 n ^{2}}{4 n + 3 n ^{2}} + \frac{32 + n - 3 n ^{2}}{8 n + 3 n ^{2}} + \frac{48 - 3 n - 3 n ^{2}}{12 n + 3 n ^{2}} + \dots + \frac{25 n - 7 n ^{2}}{7 n ^{2}}$
Then, the value of $n \to \infty lim f (n)$ is equal to

$3 + \frac{4}{3} lo g_{e} 7$

$4 - \frac{3}{4} lo g_{e} (\frac{7}{3})$

$4 - \frac{4}{3} lo g_{e} (\frac{7}{3})$

$3 + \frac{3}{4} lo g_{e} 7$

Q. For positive integer n, define f(n)=n+4n+3n216+5n−3n2​+8n+3n232+n−3n2​+12n+3n248−3n−3n2​+…+7n225n−7n2​ Then, the value of n→∞lim​f(n) is equal to

3+34​loge​7

4−43​loge​(37​)

4−34​loge​(37​)

3+43​loge​7

Q. For positive integer $n$ , define
$f (n) = n + \frac{16 + 5 n - 3 n ^{2}}{4 n + 3 n ^{2}} + \frac{32 + n - 3 n ^{2}}{8 n + 3 n ^{2}} + \frac{48 - 3 n - 3 n ^{2}}{12 n + 3 n ^{2}} + \dots + \frac{25 n - 7 n ^{2}}{7 n ^{2}}$
Then, the value of $n \to \infty lim f (n)$ is equal to

$3 + \frac{4}{3} lo g_{e} 7$

$4 - \frac{3}{4} lo g_{e} (\frac{7}{3})$

$4 - \frac{4}{3} lo g_{e} (\frac{7}{3})$

$3 + \frac{3}{4} lo g_{e} 7$