Fighter plane's analysis

Algebra Level 5

A fighter plane can travel $d_n$ kilometers in the $n$ -th hour. The values $d_n$ are related by $d_1 = 1$ and $d_{n+1} = 2d_n + n(1+2^n)$ for all $1 \le n \le 9$ . If $d_k = (k^2 - 2k + 13) \cdot 2^{k-2} - k - 1$ , determine the value of $k$ .

The answer is 7.

1 solution

Anchal Shukla
Jan 12, 2016

d (n+1) = 2dn +n + n.2^n Dividing by 2^(n+1)

d(n+1)/2^(n+1) = dn/2^n+ + n/2^(n+1) + n/2

Adding it from n=1 to n=k,

We get,

dk/2^k = (d1)/4 + (1/2^2 + 2/2^3 + 3/2^4 +..... (k-1)/2^k) + (1/2 +2/2 + 3/2 ...... + (k-1)/2)

Doing summation of brackets again and placing value of d1,

dk/2^k = 1/4 + 1/2 - (k-1)/2^k + (k-1)(k-2)/4

Now placing the value of dk from other given equation and simplifying,

2 + 2 + k^2 - 3k + 2= k^2 - 2k + 13

k = 7

Nice problem! took me some time to solve it! .is it original?

Prakhar Bindal - 4 years, 12 months ago

Solution upvoted

vineet golcha - 4 years, 11 months ago

Are bhai Slack par Aa Ja!

Prakhar Bindal - 4 years, 11 months ago

Fighter plane's analysis

The answer is 7.

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