Binomial Theorem Question - None

Question

Let $S_1 = \sum_{j=1}^{10} j(j-1)\,{}^{10}C_j$, $S_2 = \sum_{j=1}^{10} j\,{}^{10}C_j$, and $S_3 = \sum_{j=1}^{10} j^2\,{}^{10}C_j$.

Statement 1: $S_3 = 55 \times 2^9$.

Statement 2: $S_1 = 90 \times 2^8$ and $S_2 = 10 \times 2^8$.

Accepted Answer

Step 1: Find $S_2 = \sum_{j=1}^{10} j\binom{10}{j}$

Start with $(1+x)^{10} = \sum_{j=0}^{10} \binom{10}{j}x^j$

Differentiate both sides: $10(1+x)^9 = \sum_{j=1}^{10} j\binom{10}{j}x^{j-1}$

Set $x=1$: $10 \cdot 2^9 = \sum_{j=1}^{10} j\binom{10}{j}$

Therefore: $S_2 = 10 \cdot 2^9 = 10 \cdot 512 = 5120$

Note: $10 imes 2^9 eq 10 imes 2^8$ (Statement 2 claims $S_2 = 10 imes 2^8 = 1280$, which is FALSE)

Step 2: Find $S_1 = \sum_{j=1}^{10} j(j-1)\binom{10}{j}$

Differentiate $10(1+x)^9 = \sum_{j=1}^{10} j\binom{10}{j}x^{j-1}$ again:

$90(1+x)^8 = \sum_{j=2}^{10} j(j-1)\binom{10}{j}x^{j-2}$

Set $x=1$: $90 \cdot 2^8 = S_1$

Therefore: $S_1 = 90 imes 2^8$ ✓ (This part of Statement 2 is TRUE)

Step 3: Find $S_3 = \sum_{j=1}^{10} j^2\binom{10}{j}$

Note that: $j^2\binom{10}{j} = j(j-1)\binom{10}{j} + j\binom{10}{j}$

Therefore: $S_3 = S_1 + S_2 = 90 imes 2^8 + 10 imes 2^9$

$S_3 = 90 imes 2^8 + 10 imes 2 imes 2^8 = 90 imes 2^8 + 20 imes 2^8$

$S_3 = 110 imes 2^8 = 55 imes 2 imes 2^8 = 55 imes 2^9$ ✓

Therefore: Statement 1 is TRUE

Step 4: Verify Statement 2

Statement 2 claims: $S_1 = 90 imes 2^8$ (TRUE) AND $S_2 = 10 imes 2^8$ (FALSE)

Since one part is false, Statement 2 is FALSE overall.

Conclusion: Statement 1 is TRUE, Statement 2 is FALSE.

∴ Answer: D

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