Diketahui S(n)S\left(n\right)S(n) adalah rumus dari
2+8+32+⋯+4n2=23(4n−1)2+8+32+\dots+\frac{4^n}{2}=\frac{2}{3}(4^n-1)2+8+32+⋯+24n=32(4n−1)
Jika diandaikan S(n)S\left(n\right)S(n) benar untuk n=9n=9n=9, maka akan dibuktikan benar bahwa ....
2+8+32+⋯+492=23(49−1)2+8+32+\dots+\frac{4^9}{2}=\frac{2}{3}(4^9-1)2+8+32+⋯+249=32(49−1)
2+8+32+⋯+4112=23(411−1)2+8+32+\dots+\frac{4^{11}}{2}=\frac{2}{3}(4^{11}-1)2+8+32+⋯+2411=32(411−1)
2+8+32+⋯+4102=23(410−1)2+8+32+\dots+\frac{4^{10}}{2}=\frac{2}{3}(4^{10}-1)2+8+32+⋯+2410=32(410−1)
492=23(49−1)\frac{4^{9}}{2}=\frac{2}{3}(4^9-1)249=32(49−1)
4102=23(410−1)\frac{4^{10}}{2}=\frac{2}{3}(4^{10}-1)2410=32(410−1)