TS ICET 23 May 2019 Shift 2 Paper

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Question : 115 of 200

Marks: +1, -0

If f (1) = 1 and f (n) =

2Σ↖{n-1}↙{k=1}

f (k), (n > 1) then

Σ↖{m}↙{k=1}

f (k) =

$3^m - 1$
$3^m$
$3^{m-2}$
$3^{m-1}$

Solution:

Given

f(1) = 1

f(2) = 2[f(1)] = 2

f(3) = 2[f(1) + f(2)] = 2.3

f(4) = 2[f(1) + f(2) + f(3)] = 2.3^2

Similarly

f(5) = 2.33 ; f(m) = 2.3^{m – 2}

.
.

∑_{k=1}↖{m} f(k)=f(1)+f(2)+f(3) . . . . . . . f(m)

=1+2+2.3+2.3^{2}+. . . . . .+2.3^{m-2}

=1+2(1+3+3^{2}+. . . .3^{m-2})

=1+2 . \;{3^{m-1}}/{3-1}=1+3^{m-1}-1=3^{m-1}

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