Let S_1 = _j=1^10 j(j-1) 10j , S_2 = _j=1^10 j 10j and S_3 = _j=1^10 j^2 10j .Statement-1: S_3 = 55 × 2^9 Statement-2: S_1 = 90 × 2^8 and S_2 = 10 × 2 - Mathematics MCQ AIEEE 2026 | ExamTest.live

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Let S_{1} = \sum_{j = 1}^{10} j (j - 1) (j 10), S_{2} = \sum_{j = 1}^{10} j (j 10) and S_{3} = \sum_{j = 1}^{10} j^{2} (j 10) .Statement-1: S_{3} = 55 \times 2^{9} Statement-2: S_{1} = 90 \times 2^{8} and S_{2} = 10 \times 2^{8} .

A $Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation for Statement-1.$
B $Statement-1 is true, Statement-2 is false.$
C $Statement-1 is false, Statement-2 is true.$
D $Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation for Statement-1.$

Solution & Step-by-step Explanation

Using properties of binomial coefficients: S_{2} = \sum j (j n) = n 2^{n - 1} = 10 \cdot 2^{9} . (Statement 2 says 10 \cdot 2^{8}, which is False). S_{1} = \sum j (j - 1) (j n) = n (n - 1) 2^{n - 2} = 10 (9) \cdot 2^{8} = 90 \cdot 2^{8} . S_{3} = \sum j^{2} (j n) = \sum [j (j - 1) + j] (j n) = S_{1} + S_{2} = 90 \cdot 2^{8} + 10 \cdot 2^{9} = 90 \cdot 2^{8} + 20 \cdot 2^{8} = 110 \cdot 2^{8} = 55 \cdot 2^{9} .Statement-1 is True. Statement-2 is False because of S_{2} .

Try it yourself before checking the explanation above.

Let S_{1} = \sum_{j = 1}^{10} j (j - 1) (j 10), S_{2} = \sum_{j = 1}^{10} j (j 10) and S_{3} = \sum_{j = 1}^{10} j^{2} (j 10) .Statement-1: S_{3} = 55 \times 2^{9} Statement-2: S_{1} = 90 \times 2^{8} and S_{2} = 10 \times 2^{8} .

Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation for Statement-1.

Statement-1 is true, Statement-2 is false.

Statement-1 is false, Statement-2 is true.

Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation for Statement-1.

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