Q.

For k∈ℕ, if the sum of the series 1+4k+8k2+13k3+19k4+... is 10, then the value of k is __________ .             [2023]


Ans.

(2)

Given, 10=1+4k+8k2+13k3+19k4+… up to ∞

⇒9=4k+8k2+13k3+19k4+… up to ∞                    ...(i)

⇒9k=4k2+8k3+13k4+19k5+… up to ∞               ...(ii)

Subtract (ii) from (i), we get

S=9(1-1k)=4k+4k2+5k3+6k4+… up to ∞         ...(iii)

Sk=4k2+4k3+5k4+6k5+… up to ∞                          ...(iv)

Subtract (iv) from (iii), we get

(1-1k)S=4k+1k3+1k4+1k5+… up to ∞                ...(v)

Putting S=9(1-1k) from (iii) in (v), we get

          9(1-1k)2=4k+1k31-1k

⇒9(1-1k)3=4k(1-1k)+1k3⇒9(k-1)3k3=4k(k-1k)+1k3

⇒9(k-1)3=4k(k-1)+1

Put k-1=x, then

9x3=4(x+1)x+1⇒9x3=4x2+4x+1

⇒9x3-4x2-4x-1=0

⇒(x-1)(9x2+5x+1)=0⇒(x-1)=0⇒x=1

∴   k-1=1⇒k=2