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JEE Main · 2024 · 06 Apr 2024 · Shift 2 · Sequences and Series

S (x) = (1 + x) + 2 (1 + x)^{2} + 3 (1 + x)^{3} + 3 (1 + x)^{4} + \dots + 60 (1 + x)^{60}, x \neq = 0

and

(60)^{2} S (60) = a (b)^{b} + b, a, b \in N

then

(a + b)

is equal to ______.

Answer: 3660

Solution

With

r = 1 + x

,

S (x) = \sum_{k = 1}^{60} k r^{k}

. Then

(1 + x) S (x) - S (x) = \sum_{k = 1}^{60} r^{k} - 60 r^{61} = \frac{r ( r ^{60} - 1 )}{r - 1} - 60 r^{61}

, i.e.

x^{2} S (x) = - r (r^{60} - 1) + 60 x r^{61}

At

x = 60

(

r = 61

):

(60)^{2} S (60) = - 61 (6 1^{60} - 1) + 3600 \cdot 6 1^{61} = 61 + 3599 \cdot 6 1^{61}

. Matching

a b^{b} + b

with

b = 61

gives

a = 3599

, so

a + b = 3660

.

JEE Main · 2024 · 06 Apr 2024 · Shift 2 · Integer answer

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