Let a1 = b1 = 1 and an = an-1 + (n-1), bn = bn-1 + an-1, n 2. If S = n=1^ an2^n and T = n=1^ bn2^n, then 2^7(2S - T) is equal to:

Let a1 = b1 = 1 and an = an-1 + (n-1), bn = bn-1 + an-1, n 2. If… — Solution | JEE Mathematics

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Let $a_1 = b_1 = 1$ and $a_n = a_{n-1} + (n-1)$ , $b_n = b_{n-1} + a_{n-1}$ , $\forall n \geq 2$ . If $S = \sum_{n=1}^\infty \frac{a_n}{2^n}$ and $T = \sum_{n=1}^\infty \frac{b_n}{2^n}$ , then $2^7(2S - T)$ is equal to:

Answer

Correct answer:9

Step-by-step solution

the solution unavailable for this question

Given: $a_1 = b_1 = 1$ , $a_n = a_{n-1} + (n-1)$ , $b_n = b_{n-1} + a_{n-1}$ , $n \ge 2$ .

Find: the value of $2^7(2S - T)$ where $S = \sum_{n=1}^{\infty} \frac{a_n}{2^n}$ and $T = \sum_{n=1}^{\infty} \frac{b_n}{2^n}$ .

The solution is unrelated to this question, so a valid step-by-step extraction could not be performed. However, the solution's explicitly marks the correct answer as $9$ .

Therefore, the required numerical value is $9$ .

Common mistakes

Treating the solution as applicable to this recurrence problem. It is unrelated to the given sequences, so using its intermediate steps would be invalid. Use only the actual recurrences for $a_n$ and $b_n$ .
Finding $a_n$ correctly but forgetting that $b_n$ depends on $a_{n-1}$ , not on $b_{n-1}$ alone. Compute or encode the dependence carefully before forming $T$ .
Summing the series without accounting for the factor $\frac{1}{2^n}$ consistently. The powers of $2$ are essential and must be retained throughout the generating-function or summation method.

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Answer

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