Question 1083377
Using the rule of Newton Binomial:

(1+2x)^7 = sum((7Ck)(1)^(7-k)(2x)^k from k= 0 to k = 7.

Se the (1)^k = 1 for any value ok k.

= 7C0 + 7C1(2x) + 7C2(4x^2) + 7C3(8x^3) + 7C4(16x^4) + 7C5(32x^5) + 7C6(64x^6) +7C7(128x^7)

= 1 + 14x + 84x^2 + 280x^3 + 560x^4 + 672x^5 + 448x^6 + 128x^7.

Remember that nCk = n!/((n-k)!(k!)) is n>=k and n and k are natural numbers  or zero.

@natolino_