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EXERCISE 2.6 · Q91

Q.Find the sum 1×3×5+3×5×7+5×7×9+⋯+(2n−1)(2n+1)(2n+3)1\times3\times5+3\times5\times7+5\times7\times9+\cdots+(2n-1)(2n+1)(2n+3).

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(2r−1)(2r+1)=4r2−1(2r-1)(2r+1)=4r^2-1, times (2r+3)(2r+3) gives 8r3+12r2−2r−38r^3+12r^2-2r-3. Sn=8∑r3+12∑r2−2∑r−3n=8[n(n+1)2]2+12⋅n(n+1)(2n+1)6−2⋅n(n+1)2−3nS_n=8\sum r^3+12\sum r^2-2\sum r-3n=8\left[\dfrac{n(n+1)}2\right]^2+12\cdot\dfrac{n(n+1)(2n+1)}6-2\cdot\dfrac{n(n+1)}2-3n. Expanding and simplifying gives $S_n=n(2n^3+8 …

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