If displaystyle∑r=1n r4=I(n) then displaystyle∑r=1n(2 r-1)4 is equal to

Question

If displaystyle∑r=1n r4=I(n) then displaystyle∑r=1n(2 r-1)4 is equal to (A) I(2 n)-I(n) (B) I(2 n)-16 I(n) (C) I(2 n)-8 I(n) (D) I(2 n)-4 I(n)

Tardigrade Admin · Accepted Answer

I(2 n)=14+24+34+ ldots+(2 n-1)4+(2 n)4 =(14+34+54+ ldots+(2 n-1)4) +24(14+24+34+44+ ldots n4) = displaystyle∑r=1n(2 r-1)4+16 ⋅ I(n) ⇒ displaystyle∑r=1n(2 r-1)4=I(2 n)-16 I(n)

Q. If $r = 1 \sum n r^{4} = I (n)$ then $r = 1 \sum n (2 r - 1)^{4}$ is equal to

$I (2 n) - I (n)$

$I (2 n) - 16 I (n)$

$I (2 n) - 8 I (n)$

$I (2 n) - 4 I (n)$

$I (2 n) = 1^{4} + 2^{4} + 3^{4} + \dots + (2 n - 1)^{4} + (2 n)^{4}$
$= (1^{4} + 3^{4} + 5^{4} + \dots + (2 n - 1)^{4}) + 2^{4} (1^{4} + 2^{4} + 3^{4} + 4^{4} + \dots n^{4})$
$= r = 1 \sum n (2 r - 1)^{4} + 16 \cdot I (n)$
$\Rightarrow r = 1 \sum n (2 r - 1)^{4} = I (2 n) - 16 I (n)$

Q. If r=1∑n​r4=I(n) then r=1∑n​(2r−1)4 is equal to

I(2n)−I(n)

I(2n)−16I(n)

I(2n)−8I(n)

I(2n)−4I(n)